JEE Mathematics Coordinates and Straight Lines MCQs Set 14

Answer: (b) \( (2+\sqrt3,\ -2\sqrt3+1) \)

Explanation:
1. A clockwise rotation means \( \theta=-60^\circ \)
2. \( x=x'\cos\theta-y'\sin\theta=4\cdot\frac12-2\left(-\frac{\sqrt3}2\right)=2+\sqrt3 \)
3. \( y=x'\sin\theta+y'\cos\theta=4\left(-\frac{\sqrt3}2\right)+2\cdot\frac12=1-2\sqrt3 \)

Teacher's Note:
1. Clockwise rotation of axes uses a negative angle.
2. Substitute into the new-to-old formulas.

Answer: (c) (2, -3)

Explanation:
1. Put \( x=X+h \), \( y=Y+k \)
2. The linear terms are \( (2h-4)X+(2k+6)Y \)
3. Setting them to zero gives \( h=2 \), \( k=-3 \) (the centre of the circle)

Teacher's Note:
1. Expand after substituting x = X + h and y = Y + k.
2. The new origin is the centre of the circle.

Answer: (b) \( 12x^2-4y^2=3 \)

Explanation:
1. A constant difference of focal distances gives a hyperbola with foci \( (\pm1,0) \)
2. \( 2a=1\Rightarrow a=\frac12 \) and \( c=1 \), so \( b^2=c^2-a^2=\frac34 \)
3. \( \frac{x^2}{1/4}-\frac{y^2}{3/4}=1\Rightarrow12x^2-4y^2=3 \)

Teacher's Note:
1. The difference of distances equals 2a for a hyperbola.
2. Use b squared = c squared minus a squared.

Answer: (a) \( \frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1 \)

Explanation:
1. A constant sum of focal distances gives an ellipse with semi-major axis a and focal distance \( c=ae \)
2. \( b^2=a^2-c^2=a^2(1-e^2) \)
3. Equation: \( \frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1 \)

Teacher's Note:
1. Constant sum means ellipse.
2. The major axis lies along the x-axis since the foci are on it.

Answer: (b) \( 2ax+k^2=0 \)

Explanation:
1. \( PA^2-PB^2=(x-a)^2-(x+a)^2=-4ax \)
2. \( -4ax=2k^2\Rightarrow 2ax+k^2=0 \)

Teacher's Note:
1. The y squared terms cancel.
2. Use the difference of squares in x.

Answer: (a) \( x-y=0 \)

Explanation:
1. \( PA^2=PB^2 \)
2. \( (x-a-b)^2+(y-a+b)^2=(x-a+b)^2+(y-a-b)^2 \)
3. Expanding, \( -4bx+4by=0 \)
4. So \( x-y=0 \)

Teacher's Note:
1. Equal distances give the perpendicular bisector.
2. Here the segment AB has slope -1, so the bisector is y = x.

Answer: (a) hyperbola

Explanation:
1. A constant difference of the distances from two fixed points is the definition of a hyperbola
2. The constant must be less than AB

Teacher's Note:
1. Difference gives a hyperbola; sum gives an ellipse.
2. A parabola needs a point and a line.

Answer: (d) ellipse

Explanation:
1. A constant sum of the distances from two fixed points is the definition of an ellipse
2. The constant must be greater than AB

Teacher's Note:
1. A sum gives an ellipse.
2. Remember the constant must exceed the distance between the foci.

Answer: (b) the perpendicular bisector of the line segment AB

Explanation:
1. \( (x-a)^2+y^2=(x+a)^2+y^2\Rightarrow x=0 \)
2. That is the y-axis, which is the perpendicular bisector of AB

Teacher's Note:
1. Equidistance from two points gives their perpendicular bisector.
2. Here AB lies on the x-axis, so the bisector is the y-axis.

Answer: (b) \( y=2x \)

Explanation:
1. Distance from the x-axis is \(
2. y
3. \) and from the y-axis is \(
4. x
5. \)
6. \(
7. y
8. =2
9. x
10. \)
11. The option that represents one branch is \( y=2x \) (the other branch is \( y=-2x \))

Teacher's Note:
1. The x-axis distance uses y and vice versa.
2. The locus is a pair of lines through the origin.

Answer: (b) circle

Explanation:
1. Take \( A(-a,0) \), \( B(a,0) \), \( P(x,y) \)
2. \( PA^2+PB^2=2x^2+2y^2+2a^2=c \)
3. So \( x^2+y^2 \) is constant, a circle centred at the mid-point of AB

Teacher's Note:
1. Choose axes through the mid-point of AB.
2. The cross terms cancel, leaving a circle.

Answer: (b) \( y^2+12x=36 \)

Explanation:
1. The condition is \( \sqrt{x^2+y^2}+
2. x-2
3. =4 \)
4. For \( x\ge2 \): \( \sqrt{x^2+y^2}=6-x \), so \( x^2+y^2=36-12x+x^2 \)
5. That gives \( y^2+12x=36 \)
6. (For \( x<2 \) the locus is \( y^2=4x+4 \), which is not an option)

Teacher's Note:
1. Split the modulus into two cases.
2. Square the equation carefully.

Answer: (b) \( y^2-2ax+a^2=0 \)

Explanation:
1. \( (x-a)^2+y^2=x^2 \)
2. \( -2ax+a^2+y^2=0 \)
3. So \( y^2-2ax+a^2=0 \), a parabola

Teacher's Note:
1. The distance from the y-axis is
2. x
3. .
4. Square both sides.

Answer: (c) \( x^2+y^2=a^2+b^2 \)

Explanation:
1. Square both equations and add
2. \( (x\cos\alpha+y\sin\alpha)^2+(x\sin\alpha-y\cos\alpha)^2=a^2+b^2 \)
3. The left side simplifies to \( x^2+y^2 \)
4. The locus is the circle \( x^2+y^2=a^2+b^2 \)

Teacher's Note:
1. Squaring and adding eliminates the variable angle.
2. The cross terms cancel.

Answer: (b) \( (x+3y+1)(x+3y-23)=0 \)

Explanation:
1. \( AB=\sqrt{36+4}=2\sqrt{10} \) and the line AB is \( x+3y=11 \)
2. Area \( =\frac12\cdot AB\cdot d=12\Rightarrow d=\frac{12}{\sqrt{10}} \)
3. \( \frac{
4. x+3y-11
5. }{\sqrt{10}}=\frac{12}{\sqrt{10}}\Rightarrow x+3y-11=\pm12 \)
6. The two lines are \( x+3y-23=0 \) and \( x+3y+1=0 \)

Teacher's Note:
1. The locus is a pair of lines parallel to AB.
2. Use area equals half base times the perpendicular distance.

Answer: (b) \( (3x-1)^2+(3y)^2=a^2+b^2 \)

Explanation:
1. \( 3x=a\cos t+b\sin t+1 \) and \( 3y=a\sin t-b\cos t \)
2. \( 3x-1=a\cos t+b\sin t \)
3. Squaring and adding: \( (3x-1)^2+(3y)^2=a^2+b^2 \)

Teacher's Note:
1. Eliminate the parameter by squaring and adding.
2. The cross terms cancel.

Answer: (d) \( x^2+y^2=a^2 \)

Explanation:
1. The angle in a semicircle is a right angle
2. So P lies on the circle with diameter AB, centre the origin and radius a
3. \( x^2+y^2=a^2 \) (excluding A and B themselves)

Teacher's Note:
1. Use the semicircle theorem.
2. Or use the perpendicularity of PA and PB: slopes multiply to -1.

Answer: (a) \( (x-3y)(x+3y)=0 \)

Explanation:
1. Area of POA \( =\frac12\cdot4\cdot
2. x
3. =2
4. x
5. \)
6. Area of POB \( =\frac12\cdot6\cdot
7. y
8. =3
9. y
10. \)
11. \( 2
12. x
13. =2\cdot3
14. y
15. \Rightarrow
16. x
17. =3
18. y
19. \)
20. So \( x=3y \) or \( x=-3y \), i.e. \( (x-3y)(x+3y)=0 \)

Teacher's Note:
1. Use the base on each axis and the perpendicular distance of P.
2. The modulus gives two lines.

Answer: (d) \( \frac{y^2}9-\frac{x^2}7=1 \)

Explanation:
1. The foci are \( (0,\pm4) \), so \( c=4 \) and \( 2a=6\Rightarrow a=3 \)
2. \( b^2=c^2-a^2=7 \)
3. The transverse axis is along the y-axis: \( \frac{y^2}9-\frac{x^2}7=1 \)

Teacher's Note:
1. The foci lie on the y-axis, so y squared comes first.
2. Use b squared = c squared minus a squared.

Answer: (d) a straight line parallel to the y-axis

Explanation:
1. \( (x+1)^2+y^2+(x-2)^2+y^2=2\left[(x-1)^2+y^2\right] \)
2. \( 2x^2-2x+5+2y^2=2x^2-4x+2+2y^2 \)
3. \( 2x+3=0\Rightarrow x=-\frac32 \), a line parallel to the y-axis

Teacher's Note:
1. The squared terms cancel.
2. A vertical line has the form x = constant.

Answer: (a) \( y\lambda=2a \)

Explanation:
1. Let \( C(x,y) \)
2. \( \cot A=\frac{a-x}{y} \) and \( \cot B=\frac{a+x}{y} \)
3. Sum \( =\frac{2a}y=\lambda \)
4. So \( \lambda y=2a \), a line parallel to AB

Teacher's Note:
1. Use the horizontal distance over the vertical height for each cotangent.
2. The x terms cancel.

Answer: (b) an ellipse

Explanation:
1. Take the lines as axes and the ends as \( (p,0) \), \( (0,q) \) with \( p^2+q^2=(a+b)^2 \)
2. The point dividing in the ratio \( a:b \) is \( (h,k)=\left(\frac{bp}{a+b},\frac{aq}{a+b}\right) \)
3. Eliminating p and q: \( \frac{h^2}{b^2}+\frac{k^2}{a^2}=1 \), an ellipse

Teacher's Note:
1. Write the dividing point in terms of the two ends.
2. Use the fixed length to eliminate them.