RBSE Solutions Class 11 Maths Chapter 12 Conic Section Exercise 12.4

Question 1. Find the coordinates of the points of intersection where straight line \( 4y + 3x + 6 = 0 \) cuts the parabola \( 2y^2 = 9x \).
Answer:
The given equation of the straight line is:
\( 4y + 3x + 6 = 0 \)
From this, we can express \( 3x \) as:
\( 3x = -4y - 6 \) ...(i)
The given equation of the parabola is:
\( 2y^2 = 9x \) ...(ii)
Now, we substitute the value of \( x \) from equation (i) into equation (ii). First, let's find \( x \) from (i): \( x = \frac{-4y-6}{3} \).
Substitute this \( x \) into \( 2y^2 = 9x \):
\( 2y^2 = 9 \left( \frac{-4y-6}{3} \right) \)
\( \implies 2y^2 = 3(-4y - 6) \)
\( \implies 2y^2 = -12y - 18 \)
Bring all terms to one side to form a quadratic equation:
\( \implies 2y^2 + 12y + 18 = 0 \)
Divide the entire equation by 2 to simplify:
\( \implies y^2 + 6y + 9 = 0 \)
This is a perfect square trinomial, which can be factored as \( (y+3)^2 \):
\( \implies (y+3)^2 = 0 \)
Solving for \( y \), we get:
\( \implies y = -3 \)
Now, substitute this value of \( y = -3 \) back into equation (ii) to find \( x \):
\( 2y^2 = 9x \)
\( \implies 2(-3)^2 = 9x \)
\( \implies 2(9) = 9x \)
\( \implies 18 = 9x \)
\( \implies x = \frac{18}{9} \)
\( \implies x = 2 \)
So, the coordinates of the point of intersection are \( (2, -3) \).
In simple words: We find where the straight line and the curve meet by solving their equations together. We substitute one equation into the other to get a value for 'y', then use that 'y' to find 'x'. This gives us the exact spot where they cross.

🎯 Exam Tip: Always verify your solution by plugging the intersection coordinates back into both original equations to ensure they satisfy both equations.

Question 2. Find the length of chord cut by line \( 4y - 8 \) at parabola \( y^2 = 8x \).
Answer:
The given equation of the parabola is:
\( y^2 = 8x \) ...(i)
The given equation of the line is (assuming a typo in the question and using the solution's equation for the line):
\( 4y - 3x = 8 \)
We can rearrange the line equation to express \( y \):
\( 4y = 3x + 8 \)
\( \implies y = \frac{3x+8}{4} \) ...(ii)
Now, substitute this expression for \( y \) from equation (ii) into the parabola equation (i):
\( \left( \frac{3x+8}{4} \right)^2 = 8x \)
\( \implies \frac{(3x+8)^2}{16} = 8x \)
\( \implies (3x+8)^2 = 128x \)
Expand the left side:
\( \implies 9x^2 + 48x + 64 = 128x \)
Rearrange into a standard quadratic form:
\( \implies 9x^2 + 48x - 128x + 64 = 0 \)
\( \implies 9x^2 - 80x + 64 = 0 \)
To solve this quadratic equation, we use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). Here, \( a=9 \), \( b=-80 \), \( c=64 \).
\( x = \frac{-(-80) \pm \sqrt{(-80)^2 - 4(9)(64)}}{2(9)} \)
\( \implies x = \frac{80 \pm \sqrt{6400 - 2304}}{18} \)
\( \implies x = \frac{80 \pm \sqrt{4096}}{18} \)
The square root of 4096 is 64.
\( \implies x = \frac{80 \pm 64}{18} \)
This gives us two possible values for \( x \):
\( x_1 = \frac{80 + 64}{18} = \frac{144}{18} = 8 \)
\( x_2 = \frac{80 - 64}{18} = \frac{16}{18} = \frac{8}{9} \)
Now, substitute these \( x \) values back into equation (i) to find the corresponding \( y \) values.
For \( x_1 = 8 \):
\( y^2 = 8(8) \)
\( \implies y^2 = 64 \)
\( \implies y = \pm 8 \)
From the line equation \( 4y = 3x+8 \): \( 4y = 3(8)+8 = 24+8 = 32 \implies y = 8 \). So, one point is \( (8, 8) \).
For \( x_2 = \frac{8}{9} \):
\( y^2 = 8 \left( \frac{8}{9} \right) \)
\( \implies y^2 = \frac{64}{9} \)
\( \implies y = \pm \frac{8}{3} \)
From the line equation \( 4y = 3x+8 \): \( 4y = 3\left(\frac{8}{9}\right)+8 = \frac{8}{3}+8 = \frac{8+24}{3} = \frac{32}{3} \implies y = \frac{8}{3} \). So, the other point is \( \left(\frac{8}{9}, \frac{8}{3}\right) \).
The coordinates of the ends of the chord are \( (8, 8) \) and \( \left(\frac{8}{9}, \frac{8}{3}\right) \).
Now, we find the length of the chord using the distance formula \( D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \):
\( D = \sqrt{\left(\frac{8}{9} - 8\right)^2 + \left(\frac{8}{3} - 8\right)^2} \)
\( D = \sqrt{\left(\frac{8 - 72}{9}\right)^2 + \left(\frac{8 - 24}{3}\right)^2} \)
\( D = \sqrt{\left(\frac{-64}{9}\right)^2 + \left(\frac{-16}{3}\right)^2} \)
\( D = \sqrt{\frac{4096}{81} + \frac{256}{9}} \)
To add these fractions, find a common denominator, which is 81:
\( D = \sqrt{\frac{4096}{81} + \frac{256 \times 9}{9 \times 9}} \)
\( D = \sqrt{\frac{4096 + 2304}{81}} \)
\( D = \sqrt{\frac{6400}{81}} \)
\( D = \frac{\sqrt{6400}}{\sqrt{81}} \)
\( D = \frac{80}{9} \)
Thus, the length of the chord is \( \frac{80}{9} \) units.
In simple words: First, we find the two points where the line cuts the curve by solving their equations. Then, we use the distance formula to calculate the length between these two points, which gives us the length of the chord.

🎯 Exam Tip: When finding the length of a chord, make sure to correctly identify both intersection points and then apply the distance formula precisely. Always double-check your arithmetic, especially with fractions.

Question 3. Prove that straight line \( x + y = 1 \) touches parabola \( y = x - x^2 \).
Answer:
We are given the equation of the parabola:
\( y = x - x^2 \) ...(i)
And the equation of the straight line:
\( x + y = 1 \) ...(ii)
To find the intersection points, we solve these two equations simultaneously. From equation (ii), we can write \( y = 1 - x \).
Substitute this expression for \( y \) into equation (i):
\( 1 - x = x - x^2 \)
Rearrange the terms to form a quadratic equation:
\( \implies x^2 - x - x + 1 = 0 \)
\( \implies x^2 - 2x + 1 = 0 \)
This is a quadratic equation. For a line to touch a parabola (i.e., be tangent to it), there should be exactly one point of intersection. In a quadratic equation, this happens when the discriminant \( B^2 - 4AC \) is equal to zero (meaning it has equal and coinciding roots).
Here, \( a=1 \), \( b=-2 \), and \( c=1 \). Let's calculate the discriminant:
\( \implies B^2 - 4AC = (-2)^2 - 4(1)(1) \)
\( \implies B^2 - 4AC = 4 - 4 \)
\( \implies B^2 - 4AC = 0 \)
Since the discriminant is 0, the quadratic equation has equal roots, which means the line touches the parabola at exactly one point. Therefore, the straight line \( x + y = 1 \) is tangent to the parabola \( y = x - x^2 \).
In simple words: To prove a line touches a curve, we solve their equations together. If the result is a quadratic equation that has only one answer (meaning its special discriminant number is zero), then the line just touches the curve at one point, acting as a tangent.

🎯 Exam Tip: Remember that a line touches a parabola if, when you solve their equations simultaneously, the resulting quadratic equation has a discriminant equal to zero. This is a key condition for tangency.

Question 4. Find the condition that line \( lx + my + n = 0 \) touches the parabola \( y^2 = 4ax \).
Answer:
The given equation of the parabola is:
\( y^2 = 4ax \)
The given equation of the straight line is:
\( lx + my + n = 0 \)
We need to find the condition for this line to be tangent to the parabola. First, let's rewrite the line equation in the slope-intercept form \( y = mx + c \).
\( my = -lx - n \)
\( \implies y = \left(-\frac{l}{m}\right)x - \frac{n}{m} \)
Comparing this with \( y = M_s x + C_s \), where \( M_s \) is the slope of the line and \( C_s \) is its y-intercept, we have:
\( M_s = -\frac{l}{m} \)
\( C_s = -\frac{n}{m} \)
We know that a line \( y = M_s x + C_s \) touches the parabola \( y^2 = 4ax \) if the condition \( C_s = \frac{a}{M_s} \) is satisfied.
Substitute the values of \( M_s \) and \( C_s \) into this condition:
\( -\frac{n}{m} = \frac{a}{-\frac{l}{m}} \)
\( \implies -\frac{n}{m} = -\frac{am}{l} \)
Multiply both sides by \( -ml \):
\( \implies nl = am^2 \)
\( \implies am^2 = nl \)
This is the required condition for the line \( lx + my + n = 0 \) to touch the parabola \( y^2 = 4ax \).
In simple words: To find the condition for a line to touch a parabola, we write the line's equation in a specific form and then use a known rule that connects the parabola's 'a' value with the line's slope and y-intercept. This gives us a simple equation involving the line's coefficients and 'a'.

🎯 Exam Tip: Always remember the standard condition for a line \( y = Mx + C \) to be tangent to \( y^2 = 4ax \), which is \( C = a/M \). Convert the given line equation to this form before applying the condition.

Question 5. Prove that the length of focal chord of parabola \( y^2 = 4ax \) making an angle \( \alpha \) with x-axis will be \( 4a \cosec^2 \alpha \).
Answer:
Let the focal chord of the parabola \( y^2 = 4ax \) pass through the focus \( S(a, 0) \).
Let the coordinates of the ends of the focal chord be \( P(at^2, 2at) \) and \( Q\left(a\left(\frac{1}{t}\right)^2, 2a\left(-\frac{1}{t}\right)\right) \), which simplifies to \( Q\left(\frac{a}{t^2}, -\frac{2a}{t}\right) \).
The slope of the chord PQ is given by \( m = \tan \alpha \), where \( \alpha \) is the angle the chord makes with the x-axis.
The slope of the line segment PQ is also given by:
\( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-\frac{2a}{t} - 2at}{\frac{a}{t^2} - at^2} \)
\( \implies m = \frac{-2a\left(\frac{1}{t} + t\right)}{a\left(\frac{1}{t^2} - t^2\right)} \)
\( \implies m = \frac{-2a\left(\frac{1+t^2}{t}\right)}{a\left(\frac{1-t^4}{t^2}\right)} \)
\( \implies m = \frac{-2(1+t^2)}{t} \times \frac{t^2}{1-t^4} \)
\( \implies m = \frac{-2t(1+t^2)}{(1-t^2)(1+t^2)} \)
\( \implies m = \frac{-2t}{1-t^2} = \frac{2t}{t^2-1} \)
So, we have \( \tan \alpha = \frac{2t}{t^2-1} \). This is the formula for \( \tan(2\theta) \) if \( t = \tan \theta \), which means \( \alpha = 2\theta \). However, we are using \( t \) as a parameter here.
We also know that \( \sin \alpha = \frac{2t}{t^2+1} \) and \( \cos \alpha = \frac{t^2-1}{t^2+1} \).
The length of the focal chord PQ is given by the distance formula between \( P(at^2, 2at) \) and \( Q\left(\frac{a}{t^2}, -\frac{2a}{t}\right) \):
\( \text{Length PQ} = \sqrt{\left(at^2 - \frac{a}{t^2}\right)^2 + \left(2at - \left(-\frac{2a}{t}\right)\right)^2} \)
\( \text{Length PQ} = \sqrt{a^2\left(t^2 - \frac{1}{t^2}\right)^2 + 4a^2\left(t + \frac{1}{t}\right)^2} \)
\( \text{Length PQ} = \sqrt{a^2\left(\frac{t^4-1}{t^2}\right)^2 + 4a^2\left(\frac{t^2+1}{t}\right)^2} \)
\( \text{Length PQ} = \sqrt{a^2\frac{(t^2-1)^2(t^2+1)^2}{t^4} + 4a^2\frac{(t^2+1)^2}{t^2}} \)
Factor out common terms \( a^2 \frac{(t^2+1)^2}{t^4} \):
\( \text{Length PQ} = \sqrt{a^2 \frac{(t^2+1)^2}{t^4} \left((t^2-1)^2 + 4t^2\right)} \)
\( \text{Length PQ} = \frac{a(t^2+1)}{t^2} \sqrt{t^4 - 2t^2 + 1 + 4t^2} \)
\( \text{Length PQ} = \frac{a(t^2+1)}{t^2} \sqrt{t^4 + 2t^2 + 1} \)
\( \text{Length PQ} = \frac{a(t^2+1)}{t^2} \sqrt{(t^2+1)^2} \)
\( \text{Length PQ} = \frac{a(t^2+1)}{t^2} (t^2+1) \)
\( \text{Length PQ} = a \frac{(t^2+1)^2}{t^2} \)
We know that \( \sin \alpha = \frac{2t}{t^2+1} \), so \( \frac{1}{\sin \alpha} = \frac{t^2+1}{2t} \).
Therefore, \( \left(\frac{1}{\sin \alpha}\right)^2 = \left(\frac{t^2+1}{2t}\right)^2 = \frac{(t^2+1)^2}{4t^2} \).
So, \( \frac{(t^2+1)^2}{t^2} = 4 \left(\frac{1}{\sin \alpha}\right)^2 = 4 \cosec^2 \alpha \).
Substitute this back into the length formula:
\( \text{Length PQ} = a \times 4 \cosec^2 \alpha \)
\( \implies \text{Length PQ} = 4a \cosec^2 \alpha \)
This proves the required length of the focal chord.
In simple words: We find the length of a chord that passes through the special 'focus' point of the parabola. We use a parameter 't' to define the endpoints of this chord. After using the distance formula and simplifying the expression, we relate the parameter 't' to the angle 'alpha' the chord makes with the x-axis using trigonometric identities. This leads to the final formula for the chord's length.

🎯 Exam Tip: When dealing with focal chords of a parabola, it's often helpful to use parametric coordinates \( (at^2, 2at) \) for the points on the parabola. Remembering the relationship between \( t \) and trigonometric functions like \( \tan \alpha = \frac{2t}{1-t^2} \) or \( \sin \alpha = \frac{2t}{1+t^2} \) is crucial for these proofs.

Question 6. Find the condition that line \( x \cos \alpha + y \sin \alpha = P \) touches the parabola \( y^2 = 4ax \).
Answer:
The equation of the parabola is:
\( y^2 = 4ax \)
The equation of the straight line is:
\( x \cos \alpha + y \sin \alpha = P \)
To find the tangency condition, we convert the line equation into the slope-intercept form \( y = mx + c \).
\( y \sin \alpha = -x \cos \alpha + P \)
\( \implies y = -\frac{\cos \alpha}{\sin \alpha} x + \frac{P}{\sin \alpha} \)
\( \implies y = -(\cot \alpha) x + \frac{P}{\sin \alpha} \)
Comparing this with \( y = mx + c \), we identify:
\( m = -\cot \alpha \)
\( c = \frac{P}{\sin \alpha} \)
For the line \( y = mx + c \) to touch the parabola \( y^2 = 4ax \), the condition is \( c = \frac{a}{m} \).
Substitute the expressions for \( m \) and \( c \) into this condition:
\( \frac{P}{\sin \alpha} = \frac{a}{-\cot \alpha} \)
\( \implies \frac{P}{\sin \alpha} = \frac{a}{-\frac{\cos \alpha}{\sin \alpha}} \)
\( \implies \frac{P}{\sin \alpha} = -\frac{a \sin \alpha}{\cos \alpha} \)
Multiply both sides by \( \sin \alpha \):
\( P = -\frac{a \sin^2 \alpha}{\cos \alpha} \)
This can also be written as:
\( P \cos \alpha = -a \sin^2 \alpha \)
The required condition is \( P \cos \alpha + a \sin^2 \alpha = 0 \).
In simple words: We want to find the rule that makes a given line just touch a parabola. We change the line's equation into a standard form to find its slope and y-intercept. Then, we use a known mathematical condition for tangency (y-intercept equals 'a' divided by slope) and substitute our values. This gives us the special relationship needed for the line to be a tangent.

🎯 Exam Tip: When dealing with trigonometric forms of line equations, remember to convert them to the standard \( y = mx + c \) form before applying the tangency conditions for parabolas.

Question 7. Find the equation of tangents at the following parabolas:
(i) \( y^2 = 6x \), which is parallel to line \( 2x - 3y = 4 \)
(ii) \( y^2 = 8x \), which is perpendicular to line \( 2x - y + 1 = 0 \)
Answer:
(i) Equation of parabola: \( y^2 = 6x \)
Comparing with \( y^2 = 4ax \), we get \( 4a = 6 \), so \( a = \frac{6}{4} = \frac{3}{2} \).
Equation of straight line: \( 2x - 3y = 4 \)
Rearranging into slope-intercept form \( y = mx + c \):
\( 3y = 2x - 4 \)
\( y = \frac{2}{3}x - \frac{4}{3} \)
The slope of this line is \( m_1 = \frac{2}{3} \).
Since the tangent is parallel to this line, its slope will also be \( m = m_1 = \frac{2}{3} \).
The equation of a tangent to \( y^2 = 4ax \) with slope \( m \) is \( y = mx + \frac{a}{m} \).
Substitute \( a = \frac{3}{2} \) and \( m = \frac{2}{3} \):
\( y = \frac{2}{3}x + \frac{\frac{3}{2}}{\frac{2}{3}} \)
\( \implies y = \frac{2}{3}x + \frac{3}{2} \times \frac{3}{2} \)
\( \implies y = \frac{2}{3}x + \frac{9}{4} \)
To remove fractions, multiply the entire equation by 12 (the LCM of 3 and 4):
\( 12y = 12\left(\frac{2}{3}x\right) + 12\left(\frac{9}{4}\right) \)
\( \implies 12y = 8x + 27 \)
Rearrange to the general form:
\( \implies 8x - 12y + 27 = 0 \)
This is the required equation of the tangent.

(ii) Equation of parabola: \( y^2 = 8x \)
Comparing with \( y^2 = 4ax \), we get \( 4a = 8 \), so \( a = 2 \).
Equation of straight line: \( 2x - y + 1 = 0 \)
Rearranging into slope-intercept form \( y = mx + c \):
\( y = 2x + 1 \)
The slope of this line is \( m_1 = 2 \).
Since the tangent is perpendicular to this line, its slope \( m \) will be the negative reciprocal of \( m_1 \).
\( m = -\frac{1}{m_1} = -\frac{1}{2} \).
The equation of a tangent to \( y^2 = 4ax \) with slope \( m \) is \( y = mx + \frac{a}{m} \).
Substitute \( a = 2 \) and \( m = -\frac{1}{2} \):
\( y = -\frac{1}{2}x + \frac{2}{-\frac{1}{2}} \)
\( \implies y = -\frac{1}{2}x - 4 \)
To remove fractions, multiply the entire equation by 2:
\( 2y = -x - 8 \)
Rearrange to the general form:
\( \implies x + 2y + 8 = 0 \)
This is the required equation of the tangent.
In simple words: To find the equation of a tangent line, first find the value of 'a' from the parabola's equation. Then, determine the tangent's slope based on whether it's parallel or perpendicular to the given line. Finally, use the standard tangent formula \( y = mx + \frac{a}{m} \) to get the tangent equation.

🎯 Exam Tip: Remember the general equation of a tangent to \( y^2 = 4ax \) with slope \( m \) is \( y = mx + a/m \). For parallel lines, slopes are equal. For perpendicular lines, slopes are negative reciprocals of each other (i.e., \( m_1 m_2 = -1 \)).

Question 8. For which value of \( k \) line \( 2x - 3y - k \) will touch parabola \( y^2 = 6x \)?
Answer:
The equation of the parabola is:
\( y^2 = 6x \)
Comparing this with \( y^2 = 4ax \), we get \( 4a = 6 \), which means \( a = \frac{6}{4} = \frac{3}{2} \).
The equation of the line is:
\( 2x - 3y - k = 0 \)
We need to find the value of \( k \) for which this line touches the parabola. First, convert the line equation to the slope-intercept form \( y = mx + c \):
\( 3y = 2x - k \)
\( \implies y = \frac{2}{3}x - \frac{k}{3} \)
Comparing this with \( y = mx + c \), we find:
\( m = \frac{2}{3} \)
\( c = -\frac{k}{3} \)
For a line to touch the parabola \( y^2 = 4ax \), the condition is \( c = \frac{a}{m} \).
Substitute the values of \( a \), \( m \), and \( c \) into this condition:
\( -\frac{k}{3} = \frac{\frac{3}{2}}{\frac{2}{3}} \)
\( \implies -\frac{k}{3} = \frac{3}{2} \times \frac{3}{2} \)
\( \implies -\frac{k}{3} = \frac{9}{4} \)
Now, solve for \( k \):
\( \implies k = -3 \times \frac{9}{4} \)
\( \implies k = -\frac{27}{4} \)
So, for the line to touch the parabola, the value of \( k \) must be \( -\frac{27}{4} \).
In simple words: We use the condition that a line touches a parabola: the line's y-intercept must be equal to 'a' divided by its slope. We find 'a' from the parabola, the slope and y-intercept from the line, and then solve for 'k' using this condition.

🎯 Exam Tip: Always clearly identify the 'a' value from the parabola equation and the 'm' and 'c' values from the line equation before applying the tangency condition \( c = a/m \). Be careful with algebraic signs, especially when solving for \( k \).

Question 9. Find the equation of tangents which are drawn from point \( (4, 10) \) to parabola \( y^2 = 8x \).
Answer:
The equation of the parabola is \( y^2 = 8x \). Comparing this with \( y^2 = 4ax \), we get \( 4a = 8 \), so \( a = 2 \).
The given point is \( (x_1, y_1) = (4, 10) \).
The equation of a tangent to the parabola \( y^2 = 4ax \) from an external point \( (x_1, y_1) \) is found using the condition \( SS_1 = T^2 \), where:
\( S = y^2 - 4ax \)
\( S_1 = y_1^2 - 4ax_1 \)
\( T = yy_1 - 2a(x + x_1) \)
Let's calculate each part:
\( S = y^2 - 8x \)
\( S_1 = (10)^2 - 8(4) = 100 - 32 = 68 \)
\( T = y(10) - 2(2)(x + 4) = 10y - 4(x + 4) = 10y - 4x - 16 \)
Now, apply the condition \( SS_1 = T^2 \):
\( (y^2 - 8x)(68) = (10y - 4x - 16)^2 \)
\( 68y^2 - 544x = (10y - (4x + 16))^2 \)
\( 68y^2 - 544x = (10y)^2 + (4x + 16)^2 - 2(10y)(4x + 16) \)
\( 68y^2 - 544x = 100y^2 + (16x^2 + 128x + 256) - 80xy - 320y \)
Rearrange all terms to one side:
\( 0 = 100y^2 - 68y^2 + 16x^2 + 128x + 544x + 256 - 80xy - 320y \)
\( 0 = 32y^2 + 16x^2 + 672x + 256 - 80xy - 320y \)
Divide the entire equation by 16 to simplify:
\( 0 = 2y^2 + x^2 + 42x + 16 - 5xy - 20y \)
Rearranging the terms in a standard form:
\( x^2 + 2y^2 - 5xy + 42x - 20y + 16 = 0 \)
This is the combined equation of the two tangents drawn from the point \( (4, 10) \) to the parabola.
In simple words: To find the equation of tangents from an outside point to a parabola, we use a special formula that combines the parabola's equation, the point's coordinates, and the tangent's equation. This formula helps us find a single equation that represents both tangent lines together.

🎯 Exam Tip: The combined equation of tangents from an external point \( (x_1, y_1) \) to a parabola \( y^2 = 4ax \) is \( SS_1 = T^2 \). This is a crucial formula for such problems. Ensure careful expansion and algebraic manipulation, especially with squaring binomials.

Question 10. Find the equation of normal at the following parabolas:
(i) At point \( (2, 4) \) on \( y^2 = 8x \)
(ii) Upper side of latus rectum of \( y^2 + 12x = 0 \)
Answer:
(i) Equation of parabola: \( y^2 = 8x \)
Comparing with \( y^2 = 4ax \), we get \( 4a = 8 \), so \( a = 2 \).
The given point is \( (x_1, y_1) = (2, 4) \).
The slope of the tangent at \( (x_1, y_1) \) to \( y^2 = 4ax \) is \( m_t = \frac{2a}{y_1} \).
\( m_t = \frac{2(2)}{4} = \frac{4}{4} = 1 \).
The normal is perpendicular to the tangent, so its slope \( m_n = -\frac{1}{m_t} = -\frac{1}{1} = -1 \).
Now, use the point-slope form of a line \( y - y_1 = m_n(x - x_1) \):
\( y - 4 = -1(x - 2) \)
\( y - 4 = -x + 2 \)
Rearrange into the general form:
\( x + y - 4 - 2 = 0 \)
\( x + y - 6 = 0 \)
This is the required equation of the normal at point \( (2, 4) \).

(ii) Equation of parabola: \( y^2 + 12x = 0 \)
This can be written as \( y^2 = -12x \).
Comparing with \( y^2 = -4ax \), we get \( -4a = -12 \), so \( a = 3 \).
The latus rectum for \( y^2 = -4ax \) is the line \( x = -a \), which is \( x = -3 \).
The endpoints of the latus rectum are \( (-a, \pm 2a) \). So, \( (-3, \pm 2(3)) \), which are \( (-3, 6) \) and \( (-3, -6) \).
We need the normal at the "upper side of latus rectum", which means the point \( (-3, 6) \).
So, \( (x_1, y_1) = (-3, 6) \).
The slope of the tangent at \( (x_1, y_1) \) to \( y^2 = -4ax \) is \( m_t = -\frac{2a}{y_1} \).
\( m_t = -\frac{2(3)}{6} = -\frac{6}{6} = -1 \).
The slope of the normal \( m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1 \).
Now, use the point-slope form \( y - y_1 = m_n(x - x_1) \):
\( y - 6 = 1(x - (-3)) \)
\( y - 6 = x + 3 \)
Rearrange into the general form:
\( x - y + 3 + 6 = 0 \)
\( x - y + 9 = 0 \)
This is the required equation of the normal at the upper end of the latus rectum.
In simple words: To find the equation of a normal line to a parabola at a specific point, first find the parabola's 'a' value. Then, calculate the slope of the tangent at that point. Since the normal is perpendicular to the tangent, its slope will be the negative reciprocal. Finally, use the point-slope formula with the given point and the normal's slope to get the normal's equation.

🎯 Exam Tip: For normals, remember the relationship between tangent and normal slopes (\( m_n = -1/m_t \)). Be careful with the signs of 'a' and 'x' for parabolas like \( y^2 = -4ax \), and correctly identify the coordinates of the specific point, such as the end of the latus rectum.

Question 11. Find the equation of normal at following parabolas:
(i) \( y^2 = 4x \) which is parallel to line \( y - 2x + 5 = 0 \)
(ii) \( y^2 = 4x \) which is perpendicular to line \( x + 3y - 1 = 0 \)
Answer:
(i) Equation of parabola: \( y^2 = 4x \)
Comparing with \( y^2 = 4ax \), we get \( 4a = 4 \), so \( a = 1 \).
Equation of line: \( y - 2x + 5 = 0 \)
Rearranging into slope-intercept form: \( y = 2x - 5 \)
The slope of this line is \( m_1 = 2 \).
Since the normal is parallel to this line, its slope \( m_n \) is also \( 2 \).
The equation of a normal to \( y^2 = 4ax \) with slope \( m_n \) is \( y = m_n x - 2am_n - am_n^3 \).
Substitute \( a = 1 \) and \( m_n = 2 \):
\( y = 2x - 2(1)(2) - 1(2)^3 \)
\( y = 2x - 4 - 1(8) \)
\( y = 2x - 4 - 8 \)
\( y = 2x - 12 \)
Rearrange into the general form:
\( 2x - y - 12 = 0 \)
This is the required equation of the normal.

(ii) Equation of parabola: \( y^2 = 4x \)
Comparing with \( y^2 = 4ax \), we get \( 4a = 4 \), so \( a = 1 \).
Equation of line: \( x + 3y - 1 = 0 \)
Rearranging into slope-intercept form:
\( 3y = -x + 1 \)
\( y = -\frac{1}{3}x + \frac{1}{3} \)
The slope of this line is \( m_1 = -\frac{1}{3} \).
Since the normal is perpendicular to this line, its slope \( m_n \) will be the negative reciprocal of \( m_1 \).
\( m_n = -\frac{1}{m_1} = -\frac{1}{-\frac{1}{3}} = 3 \).
The equation of a normal to \( y^2 = 4ax \) with slope \( m_n \) is \( y = m_n x - 2am_n - am_n^3 \).
Substitute \( a = 1 \) and \( m_n = 3 \):
\( y = 3x - 2(1)(3) - 1(3)^3 \)
\( y = 3x - 6 - 27 \)
\( y = 3x - 33 \)
Rearrange into the general form:
\( 3x - y - 33 = 0 \)
This is the required equation of the normal.
In simple words: To find the normal's equation, first get 'a' from the parabola. Then, find the normal's slope. If parallel, it's the same as the given line's slope. If perpendicular, it's the negative reciprocal. Finally, use the general normal equation \( y = mx - 2am - am^3 \) with your values to get the answer.

🎯 Exam Tip: Remember the general equation of a normal to \( y^2 = 4ax \) with slope \( m \) is \( y = mx - 2am - am^3 \). Carefully determine the slope of the normal based on whether it's parallel or perpendicular to the given line.

Question 12. Prove that line \( 2x + y - 12a = 0 \) is normal chord at parabola \( y^2 = 4ax \) and its length is \( 5\sqrt{5}a \).
Answer:
The given equation of the parabola is \( y^2 = 4ax \).
The given equation of the line is \( 2x + y - 12a = 0 \).
We need to show this line is a normal chord and find its length.

First, let's find the slope of the given line. Rewrite it in slope-intercept form \( y = mx + c \):
\( y = -2x + 12a \)
So, the slope of the line is \( m = -2 \).
For a line to be a normal to the parabola \( y^2 = 4ax \), its equation must be of the form \( y = mx - 2am - am^3 \).
Substitute the slope \( m = -2 \) and the parameter \( a \) into this general normal equation:
\( y = (-2)x - 2a(-2) - a(-2)^3 \)
\( y = -2x + 4a - a(-8) \)
\( y = -2x + 4a + 8a \)
\( y = -2x + 12a \)
This equation matches the given line \( 2x + y - 12a = 0 \) (when rearranged). Therefore, the given line is indeed a normal to the parabola.

Next, we need to find the points of intersection of this normal line with the parabola to determine the length of the chord.
Substitute \( y = -2x + 12a \) into \( y^2 = 4ax \):
\( (-2x + 12a)^2 = 4ax \)
\( (2x - 12a)^2 = 4ax \)
\( 4x^2 - 48ax + 144a^2 = 4ax \)
Rearrange into a quadratic equation in terms of \( x \):
\( 4x^2 - 48ax - 4ax + 144a^2 = 0 \)
\( 4x^2 - 52ax + 144a^2 = 0 \)
Divide by 4 to simplify:
\( x^2 - 13ax + 36a^2 = 0 \)
Factor this quadratic equation:
We need two numbers that multiply to \( 36a^2 \) and add to \( -13a \). These are \( -9a \) and \( -4a \).
\( x^2 - 9ax - 4ax + 36a^2 = 0 \)
\( x(x - 9a) - 4a(x - 9a) = 0 \)
\( (x - 9a)(x - 4a) = 0 \)
So, the x-coordinates of the intersection points are \( x_1 = 9a \) and \( x_2 = 4a \).

Now, substitute these \( x \) values back into the parabola equation \( y^2 = 4ax \) to find the corresponding \( y \) values.
For \( x_1 = 9a \):
\( y^2 = 4a(9a) = 36a^2 \)
\( y = \pm \sqrt{36a^2} = \pm 6a \)
From the line equation \( y = -2x + 12a \), for \( x=9a \), \( y = -2(9a) + 12a = -18a + 12a = -6a \).
So, one intersection point is \( (9a, -6a) \).

For \( x_2 = 4a \):
\( y^2 = 4a(4a) = 16a^2 \)
\( y = \pm \sqrt{16a^2} = \pm 4a \)
From the line equation \( y = -2x + 12a \), for \( x=4a \), \( y = -2(4a) + 12a = -8a + 12a = 4a \).
So, the second intersection point is \( (4a, 4a) \).

The two intersection points are \( (P_1) = (4a, 4a) \) and \( (P_2) = (9a, -6a) \).
Now, calculate the length of the chord using the distance formula \( D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \):
\( D = \sqrt{(9a - 4a)^2 + (-6a - 4a)^2} \)
\( D = \sqrt{(5a)^2 + (-10a)^2} \)
\( D = \sqrt{25a^2 + 100a^2} \)
\( D = \sqrt{125a^2} \)
\( D = \sqrt{25 \times 5 \times a^2} \)
\( D = 5a\sqrt{5} \)
Thus, the length of the normal chord is \( 5\sqrt{5}a \) units. This matches the required length.
In simple words: First, we prove the given line is a normal to the parabola by checking if its equation fits the general form of a parabola's normal. Then, we find where this normal line crosses the parabola by solving their equations together. This gives us two points. Finally, we use the distance formula to calculate the length between these two points, which is the length of the normal chord.

🎯 Exam Tip: To prove a line is a normal, ensure its equation matches the general normal equation for the given parabola. When finding chord length, carefully solve for both intersection points and then apply the distance formula correctly.