RBSE Solutions Class 11 Maths Chapter 12 शांकव परिच्छेद More Questions

Question 1. वृत्त \( 9x^2 + y^2 + 5x = (x^2-y^2) \) की त्रिज्या है-
(A) 1
(B) 2
(C) \( \frac {4}{5} \)
(D) \( \frac {5}{4} \)
Answer: (C) \( \frac {4}{5} \)
In simple words: The radius of the given circle is \( \frac {4}{5} \). This value helps define the size of the circle.

🎯 Exam Tip: Remember that the radius of a circle must always be a positive value. Carefully identify the correct form of the equation before determining the radius.

Question 2. उस वृत्त को समीकरण जिसका केन्द्र प्रथम पाद में \( (a, \beta) \) है। तथा x-अक्ष को स्पर्श करता है, होगा-
(A) \( x^2 + y^2 – 2ax – 2\beta y + a^2 = 0 \)
(B) \( x^2 + y^2 + 2ax – 2\beta y + a^2 = 0 \)
(C) \( x^2 + y^2 – 2ax + 2\beta y + a^2 = 0 \)
(D) \( x^2 + y^2 + 2ax + 2\beta y + a^2 = 0 \)
Answer: (A) \( x^2 + y^2 – 2ax – 2\beta y + a^2 = 0 \)
In simple words: When a circle has its center at \( (a, \beta) \) and just touches the x-axis, its radius is equal to the absolute value of its y-coordinate \( \beta \). The equation reflects this by having \( \beta^2 \) cancel out.

🎯 Exam Tip: For a circle touching the x-axis, the radius is \( |\beta| \). For a circle touching the y-axis, the radius is \( |a| \). Use this fact when forming the equation.

Question 4. रेखा \( 3x + 4y = 25 \) वृत्त \( x^2 + y^2 = 25 \) को किस बिन्दु पर स्पर्श करती है
(A) (4, 3)
(B) (3, 4)
(C) (-3, -4)
(D) (3, -4)
Answer: (B) (3, 4)
In simple words: The line touches the circle at the point \( (3, 4) \). This point is on both the line and the circle.

🎯 Exam Tip: For a circle centered at the origin, the point of tangency for the line \( Ax + By = C \) is \( (Ar^2/C, Br^2/C) \).

Question 5. एक शांकवीय परिच्छेद परवलय होगा, यदि
(A) \( e = 0 \)
(B) \( e < 1 \)
(C) \( e > 1 \)
(D) \( e = 1 \)
Answer: (C) \( e > 1 \)
In simple words: The eccentricity \( e \) tells us the shape of a conic section. If \( e \) is greater than 1, the conic section is a hyperbola.

🎯 Exam Tip: The value of eccentricity \( e \) is key to identifying conic sections: \( e=0 \) for a circle, \( e<1 \) for an ellipse, \( e=1 \) for a parabola, and \( e>1 \) for a hyperbola.

Question 6. परवलय \( x^2 = -8y \) की नियता को समीकरण है
(A) \( y = -2 \)
(B) \( y = 2 \)
(C) \( x = 2 \)
(D) \( x = -2 \)
Answer: (B) \( y = 2 \)
In simple words: For a parabola that opens downwards like \( x^2 = -4ay \), its directrix is a horizontal line above it, with the equation \( y = a \). Here, \( a=2 \).

🎯 Exam Tip: Always compare the given parabola equation to its standard form (e.g., \( y^2 = 4ax \), \( x^2 = 4ay \), \( y^2 = -4ax \), or \( x^2 = -4ay \)) to correctly identify the value of 'a' and the orientation, which helps find the directrix.

Question 8. यदि किसी परवलय की नाभि (-3, 0) तथा नियता \( x + 5 = 0 \) हों तो इसका समीकरण होगा
(A) \( y^2 = 4(x + 4) \)
(B) \( y^2 + 4x + 16 = 0 \)
(C) \( y^2 + 4x = 16 \)
(D) \( x^2 = 4(y + 4) \)
Answer: (A) \( y^2 = 4(x + 4) \)
In simple words: For any point on a parabola, its distance to the focus is equal to its distance to the directrix. Using this rule, we can form the equation of the parabola.

🎯 Exam Tip: The fundamental definition of a parabola (distance to focus equals distance to directrix) is crucial for deriving its equation when given the focus and directrix.

Question 9. किसी परवलय के शीर्ष एवं नाभि क्रमशः (2, 0) तथा (5, 0) हो, तो इसका समीकरण होगा
(A) \( y^2 = 12x + 24 \)
(B) \( y^2 = 12x - 24 \)
(C) \( y^2 = -12x - 24 \)
(D) \( y^2 = -12x + 24 \)
Answer: (B) \( y^2 = 12x - 24 \)
In simple words: The vertex is at \( (2, 0) \) and the focus is at \( (5, 0) \). This means the parabola opens to the right, and the distance from the vertex to the focus is 3. Using the general form for such a parabola, we get the equation.

🎯 Exam Tip: The distance between the vertex and the focus of a parabola is always 'a'. Use the vertex and the direction it opens to select the correct standard form of the parabola equation.

Question 10. परवलय \( x^2 = -8y \) की नाभि है
(A) (2, 0)
(B) (0, 2)
(C) (-2, 0)
(D) (0, -2)
Answer: (D) (0, -2)
In simple words: For a parabola shaped like \( x^2 = -4ay \), which opens downwards, its focus is located at \( (0, -a) \). In this case, \( a \) is 2.

🎯 Exam Tip: The focus of a parabola lies on its axis of symmetry. For \( x^2 = -4ay \), the axis is the y-axis, and the parabola opens down, so the focus is at \( (0, -a) \).

Question 12. यदि रेखा \( 2y – x = 2 \) परवलय \( y^2 = 2x \) को स्पर्श करती हो, तो स्पर्श बिन्दु है
(A) (4, 3)
(B) (-4, 1)
(C) (2, 2)
(D) (1, 4)
Answer: (C) (2, 2)
In simple words: We find the point where the line meets the parabola. Since it's a tangent line, it will meet at only one point, which is \( (2, 2) \).

🎯 Exam Tip: To find the point of tangency between a line and a curve, substitute the line equation into the curve equation and solve for the common point. A tangent will result in a quadratic with a single solution.

Question 13. परवलय \( x^2 = 8y \) की रेखा \( x + 2y + 1 = 0 \) के समान्तर स्पर्श रेखा का समीकरण है-
(A) \( x + 2y + 1 = 0 \)
(B) \( x - 2y + 1 = 0 \)
(C) \( x + 2y - 1 = 0 \)
(D) \( x - 2y + 1 = 0 \)
Answer: (A) \( x + 2y + 1 = 0 \)
In simple words: We find a line that has the same slope as the given line and also touches the parabola at exactly one point. This line turns out to be \( x + 2y + 1 = 0 \).

🎯 Exam Tip: Parallel lines have the same slope. Use the slope of the given line and the tangency condition for the parabola to find the equation of the tangent.

Question 14. परवलय \( y^2 = 4x \) का एक अभिलम्ब है
(A) \( y = x + 4 \)
(B) \( y + x = 3 \)
(C) \( y + x = 2 \)
(D) \( y + x = 1 \)
Answer: (B) \( y + x = 3 \)
In simple words: A normal line is perpendicular to the tangent at the point where it touches the parabola. The equation \( y = -x + 3 \) is such a line for the parabola \( y^2 = 4x \).

🎯 Exam Tip: The equation of a normal to \( y^2 = 4ax \) with slope \( m \) is \( y = mx - 2am - am^3 \). Use this formula to test the options.

Question 15. दीर्घवृत्त \( 3x^2 + 4y^2 = 12 \) के अर्द्धनाभिलम्ब की लम्बाई होगी-
(A) \( \frac {3}{2} \)
(B) 3
(C) \( \frac {8}{\sqrt{3}} \)
(D) \( \frac {4}{\sqrt{3}} \)
Answer: (A) \( \frac {3}{2} \)
In simple words: We first rewrite the ellipse equation in standard form. Then, we use the formula \( b^2/a \) to find the length of the semi-latus rectum. This length is \( \frac {3}{2} \).

🎯 Exam Tip: Always convert the ellipse equation to its standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) to correctly identify \( a^2 \) and \( b^2 \) before calculating the semi-latus rectum.

Question 17. यदि रेखा \( y = mx + c \) दीर्घवृत्त \( \frac {{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}} =1 \) का स्पर्श करती है तो c का मान होगा-
(A) \( c = \frac {a}{m} \)
(B) \( c=\pm\sqrt{a^2m^2-b^2} \)
(C) \( c = \pm\sqrt{a^2m^2+b^2} \)
(D) \( c=a\sqrt{1+m^2} \)
Answer: (B) \( c=\pm\sqrt{a^2m^2-b^2} \)
In simple words: This formula tells us the value of the constant \( c \) for a line to touch a conic section. The square of \( c \) is equal to \( a^2m^2 \) minus \( b^2 \).

🎯 Exam Tip: Remember the condition for a line to be tangent to different conic sections. For an ellipse, the condition is usually \( c^2 = a^2m^2 + b^2 \), while for a hyperbola, it's \( c^2 = a^2m^2 - b^2 \).

Question 18. दीर्घवृत्त \( \frac {{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}} =1 \) \( (b > a) \) के नाभियों के निर्देशांक होंगे-
(A) \( (\pm ae, 0) \)
(B) \( (\pm be, 0) \)
(C) \( (0, \pm ae) \)
(D) \( (0, \pm be) \)
Answer: (D) \( (0, \pm be) \)
In simple words: When the longer axis of an ellipse is along the y-axis (meaning \( b > a \)), its foci (plural of focus) are located at points \( (0, \pm be) \). Here, \( e \) stands for eccentricity.

🎯 Exam Tip: The foci of an ellipse always lie on its major axis. If \( b > a \), the major axis is along the y-axis, and the foci will have zero x-coordinates.

Question 19. आयतीय अतिपरवलय की उत्केन्द्रता होगी
(A) 0
(B) 1
(C) \( \sqrt{2} \)
(D) 2
Answer: (C) \( \sqrt{2} \)
In simple words: A rectangular hyperbola has equal values for its 'a' and 'b' parameters. Because of this, its eccentricity (a measure of how "stretched" it is) is always \( \sqrt{2} \).

🎯 Exam Tip: For an equilateral or rectangular hyperbola, the lengths of the semi-transverse and semi-conjugate axes are equal (\( a = b \)), simplifying the eccentricity formula to \( \sqrt{2} \).

Question 21. उस वृत्त का समीकरण लिखिए जिसका केन्द्र \( (a \cos a, a \sin a) \) तथा त्रिज्या \( a \) है।
Answer: दिए गए वृत्त का केन्द्र \( (a \cos a, a \sin a) \) और त्रिज्या \( a \) है।
वृत्त का समीकरण है: \( (x - \text{केन्द्र का x-निर्देशांक})^2 + (y - \text{केन्द्र का y-निर्देशांक})^2 = (\text{त्रिज्या})^2 \)
\( \implies (x - a \cos a)^2 + (y - a \sin a)^2 = a^2 \)
\( \implies x^2 - 2ax \cos a + a^2 \cos^2 a + y^2 - 2ay \sin a + a^2 \sin^2 a = a^2 \)
\( \implies x^2 + y^2 - 2ax \cos a - 2ay \sin a + a^2(\cos^2 a + \sin^2 a) = a^2 \)
चूँकि \( \cos^2 a + \sin^2 a = 1 \) होता है,
\( \implies x^2 + y^2 - 2ax \cos a - 2ay \sin a + a^2 = a^2 \)
\( \implies x^2 + y^2 - 2ax \cos a - 2ay \sin a = 0 \)
In simple words: To find the equation of a circle, we use the formula \( (x-h)^2 + (y-k)^2 = r^2 \). We plug in the given center coordinates and the radius, then simplify. The \( a^2 \) terms cancel out, leaving \( x^2 + y^2 - 2ax \cos a - 2ay \sin a = 0 \).

🎯 Exam Tip: When given trigonometric coordinates for the center, remember the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) as it often simplifies the final equation significantly.

Question 22. यदि वृत्त \( x^2 + y^2 + 2gx + 2fy + c = 0 \) के बिन्दु \( (x_1, y_1) \) तथा \( (x_2, y_2) \) पर स्पर्श रेखाएँ परस्पर लम्बवत् हों तो सिद्ध कीजिए. \( x_1x_2 + y_1y_2 + g(x_1 + x_2) + f(y_1+y_2) + g^2 + f^2 = 0 \)
Answer: दिए गए वृत्त का समीकरण है: \( x^2 + y^2 + 2gx + 2fy + c = 0 \)
बिन्दु \( (x_1, y_1) \) पर स्पर्श रेखा का समीकरण है: \( xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0 \)
\( \implies (x_1 + g)x + (y_1 + f)y + gx_1 + fy_1 + c = 0 \)
इस रेखा का ढाल \( m_1 = -\frac{\text{x का गुणांक}}{\text{y का गुणांक}} = -\frac{(x_1+g)}{(y_1+f)} \) .....(1)
इसी तरह से, बिन्दु \( (x_2, y_2) \) पर स्पर्श रेखा का ढाल होगा:
\( m_2 = -\frac{(x_2+g)}{(y_2+f)} \)
चूँकि स्पर्श रेखाएँ परस्पर लम्बवत् हैं, तो \( m_1 \cdot m_2 = -1 \)
\( \implies \left(-\frac{(x_1+g)}{(y_1+f)}\right) \cdot \left(-\frac{(x_2+g)}{(y_2+f)}\right) = -1 \)
\( \implies \frac{(x_1+g)(x_2+g)}{(y_1+f)(y_2+f)} = -1 \)
\( \implies (x_1+g)(x_2+g) = -(y_1+f)(y_2+f) \)
\( \implies x_1x_2 + gx_1 + gx_2 + g^2 = -(y_1y_2 + fy_1 + fy_2 + f^2) \)
\( \implies x_1x_2 + gx_1 + gx_2 + g^2 = -y_1y_2 - fy_1 - fy_2 - f^2 \)
\( \implies x_1x_2 + y_1y_2 + g(x_1 + x_2) + f(y_1 + y_2) + g^2 + f^2 = 0 \)
यह सिद्ध होता है।
In simple words: We find the slopes of the tangent lines at both given points. Since the tangents are perpendicular, their slopes multiply to -1. By putting the slope formulas into this condition and simplifying, we get the required equation. This proves the relationship for perpendicular tangents.

🎯 Exam Tip: Remember the formula for the tangent to a circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \) at a point \( (x_1, y_1) \). The condition for perpendicular lines (\( m_1m_2 = -1 \)) is key to completing the proof.

Question 23. r त्रिज्या वाले उस वृत्त का समीकरण ज्ञात कीजिए जिसका केन्द्र प्रथम पाद में स्थित है तथा y-अक्ष को मूल बिन्दु से h दूरी पर स्पर्श करता है। मूल बिन्दु से होकर जाने वाली दूसरी स्पर्श रेखा का समीकरण भी ज्ञात कीजिए।
Answer: वृत्त का केन्द्र प्रथम पाद में स्थित है और y-अक्ष को मूल बिन्दु से \( h \) दूरी पर स्पर्श करता है, जिसकी त्रिज्या \( r \) है।
इसका अर्थ है कि वृत्त का केन्द्र \( (r, h) \) है और त्रिज्या \( r \) है।
वृत्त का समीकरण होगा:
\( (x - r)^2 + (y - h)^2 = r^2 \) .....(1)


माना मूल बिन्दु से होकर जाने वाली दूसरी स्पर्श रेखा का समीकरण \( y = mx \) है।
चूँकि रेखा \( y = mx \) वृत्त (1) को स्पर्श करती है, अतः वृत्त के केन्द्र \( (r, h) \) से रेखा \( y = mx \) पर डाले गए लम्ब की लम्बाई \( r \) होगी।
रेखा \( mx - y = 0 \) पर केन्द्र \( (r, h) \) से लम्ब की लम्बाई \( = \frac{|mr - h|}{\sqrt{m^2 + (-1)^2}} \)
\( \implies \frac{|mr - h|}{\sqrt{m^2 + 1}} = r \)
\( \implies |mr - h| = r\sqrt{m^2 + 1} \)
दोनों पक्षों का वर्ग करने पर:
\( \implies (mr - h)^2 = r^2(m^2 + 1) \)
\( \implies m^2r^2 - 2mrh + h^2 = m^2r^2 + r^2 \)
\( \implies -2mrh + h^2 = r^2 \)
\( \implies h^2 - r^2 = 2mrh \)
\( \implies m = \frac{h^2 - r^2}{2rh} \)
अभीष्ट स्पर्श रेखा का समीकरण होगा:
\( y = \frac{h^2 - r^2}{2rh}x \)
\( \implies 2rhy = (h^2 - r^2)x \)
\( \implies (h^2 - r^2)x - 2rhy = 0 \)
या \( (r^2 - h^2)x + 2rhy = 0 \)
In simple words: First, we find the circle's equation using its center \( (r, h) \) and radius \( r \). Then, for the tangent line through the origin, we use the formula that the perpendicular distance from the circle's center to the tangent line must equal the radius. This helps us find the slope \( m \) for the tangent, leading to its equation.

🎯 Exam Tip: Remember that if a circle touches the y-axis, its radius is the absolute value of its x-coordinate. For a tangent line passing through the origin, its equation will be of the form \( y = mx \).

Question 24. वृत्त \( x^2 + y^2 = a^2 \) के बिन्दु \( (a, \beta) \) पर खींची गई स्पर्श रेखा अक्षों को क्रमशः A एवं B बिन्दुओं पर मिलती है। सिद्ध कीजिए कि त्रिभुज OAB का क्षेत्रफल \( \frac {{a}^{4}}{2\alpha \beta } \) होगा, जहाँ O मूल बिन्दु है।
Answer: वृत्त का समीकरण है: \( x^2 + y^2 = a^2 \)
बिन्दु \( (\alpha, \beta) \) पर स्पर्श रेखा का समीकरण है: \( x\alpha + y\beta = a^2 \)
स्पर्श रेखा का x-अक्ष से अन्तःखण्ड (बिन्दु A): \( y = 0 \) रखने पर,
\( x\alpha = a^2 \)
\( \implies x = \frac{a^2}{\alpha} \)
तो, बिन्दु \( A = \left(\frac{a^2}{\alpha}, 0\right) \)
स्पर्श रेखा का y-अक्ष से अन्तःखण्ड (बिन्दु B): \( x = 0 \) रखने पर,
\( y\beta = a^2 \)
\( \implies y = \frac{a^2}{\beta} \)
तो, बिन्दु \( B = \left(0, \frac{a^2}{\beta}\right) \)
मूल बिन्दु \( O = (0, 0) \)
त्रिभुज OAB का क्षेत्रफल \( = \frac{1}{2} \times OA \times OB \)
\( = \frac{1}{2} \times \frac{a^2}{\alpha} \times \frac{a^2}{\beta} \)
\( = \frac{a^4}{2\alpha\beta} \)
यह सिद्ध होता है।
In simple words: First, we find the equation of the tangent line to the circle at the given point. Then, we find where this tangent line cuts the x-axis and y-axis to get points A and B. Finally, we use the formula for the area of a triangle \( \frac{1}{2} \times \text{base} \times \text{height} \) to calculate the area of triangle OAB. This matches the expression we needed to prove.

🎯 Exam Tip: To find the intercepts of a line, set \( y=0 \) for the x-intercept and \( x=0 \) for the y-intercept. The area of a triangle with vertices at the origin and on the axes is \( \frac{1}{2} \times |\text{x-intercept}| \times |\text{y-intercept}| \).

Question 26. परवलय \( x^2 - 4x - 8y = 4 \) की नाभि के निर्देशांक लिखिए।
Answer: दिया गया परवलय का समीकरण है: \( x^2 - 4x - 8y = 4 \)
हम इसे मानक रूप में बदलने के लिए वर्ग पूरा करेंगे।
\( \implies x^2 - 4x = 8y + 4 \)
दोनों तरफ 4 जोड़ने पर (x के वर्ग को पूरा करने के लिए):
\( \implies x^2 - 4x + 4 = 8y + 4 + 4 \)
\( \implies (x - 2)^2 = 8y + 8 \)
\( \implies (x - 2)^2 = 8(y + 1) \)
यह परवलय के मानक रूप \( X^2 = 4aY \) के समान है, जहाँ \( X = x - 2 \) और \( Y = y + 1 \)
यहां, \( 4a = 8 \implies a = 2 \)
इस प्रकार के परवलय की नाभि \( (X=0, Y=a) \) होती है।
\( \implies X = 0 \implies x - 2 = 0 \implies x = 2 \)
\( \implies Y = a \implies y + 1 = 2 \implies y = 1 \)
अतः, परवलय की नाभि के निर्देशांक \( (2, 1) \) हैं। यह नाभि अक्ष पर स्थित होती है।
In simple words: We rewrite the parabola's equation by completing the square to get it into a standard form like \( (x-h)^2 = 4a(y-k) \). From this standard form, we can easily find the 'a' value and then calculate the focus coordinates. The focus is at \( (2, 1) \).

🎯 Exam Tip: To find the focus of a parabola that is not centered at the origin, first complete the square to transform its equation into a standard form like \( (X)^2 = 4a(Y) \) or \( (Y)^2 = 4a(X) \).

Question 27. परवलय \( x^2 - 4x - 4y + 4 = 0 \) की उत्केन्द्रता लिखिए।
Answer: दिया गया परवलय का समीकरण है: \( x^2 - 4x - 4y + 4 = 0 \)
हम इसे मानक रूप में बदलने के लिए वर्ग पूरा करेंगे।
\( \implies x^2 - 4x + 4 = 4y \)
\( \implies (x - 2)^2 = 4y \)
यह परवलय के मानक रूप \( X^2 = 4aY \) के समान है, जहाँ \( X = x - 2 \) और \( Y = y \)
चूँकि यह एक परवलय है, इसकी उत्केन्द्रता (eccentricity) परिभाषा के अनुसार हमेशा 1 होती है। उत्केन्द्रता, conic section के आकार को दर्शाती है।
अतः, उत्केन्द्रता \( e = 1 \)
In simple words: We rearrange the given equation into the standard form of a parabola. By definition, a parabola always has an eccentricity of 1. So, the eccentricity of this parabola is also 1.

🎯 Exam Tip: Remember the fundamental definition of eccentricity for conic sections: \( e=0 \) for a circle, \( e<1 \) for an ellipse, \( e=1 \) for a parabola, and \( e>1 \) for a hyperbola.

Question 28. रेखा \( lx + my + n = 0 \) के परवलय \( y^2 = 4ax \) को स्पर्श करने का प्रतिबन्ध लिखिए।
Answer: दिया गया परवलय का समीकरण है: \( y^2 = 4ax \)
दिया गया रेखा का समीकरण है: \( lx + my + n = 0 \)
रेखा \( lx + my + n = 0 \) से, हम \( my = -lx - n \)
\( \implies y = \left(-\frac{l}{m}\right)x - \frac{n}{m} \)
अब, परवलय \( y^2 = 4ax \) की स्पर्श रेखा का समीकरण \( y = Mx + \frac{a}{M} \) होता है, जहाँ \( M \) स्पर्श रेखा का ढाल है।
दोनों समीकरणों की तुलना करने पर, ढाल \( M = -\frac{l}{m} \) और \( \frac{a}{M} = -\frac{n}{m} \)
\( \implies \frac{a}{-\frac{l}{m}} = -\frac{n}{m} \)
\( \implies -\frac{am}{l} = -\frac{n}{m} \)
\( \implies \frac{am}{l} = \frac{n}{m} \)
वज्रगुणन करने पर:
\( \implies am^2 = ln \)
अतः, यह शर्त है कि रेखा परवलय को स्पर्श करेगी।
In simple words: We compare the given line's equation to the general form of a tangent line for the parabola \( y^2 = 4ax \). By matching their slopes and constant terms, we find the condition \( am^2 = ln \), which ensures the line just touches the parabola.

🎯 Exam Tip: Memorize the standard condition for tangency for various conic sections. For \( y^2 = 4ax \), the line \( y = mx + c \) is tangent if \( c = a/m \), and for \( lx + my + n = 0 \), it's \( am^2 = ln \).

Question 30. परवलय \( 9y^2 – 16x - 12y - 57 = 0 \) के अक्ष का समीकरण लिखिए।
Answer: दिया गया परवलय का समीकरण है: \( 9y^2 - 16x - 12y - 57 = 0 \)
हम इसे मानक रूप में बदलने के लिए वर्ग पूरा करेंगे।
\( \implies 9y^2 - 12y = 16x + 57 \)
\( \implies 9(y^2 - \frac{12}{9}y) = 16x + 57 \)
\( \implies 9(y^2 - \frac{4}{3}y) = 16x + 57 \)
दोनों तरफ \( 9 \times \left(\frac{2}{3}\right)^2 = 9 \times \frac{4}{9} = 4 \) जोड़ने पर (y के वर्ग को पूरा करने के लिए):
\( \implies 9\left(y^2 - \frac{4}{3}y + \left(\frac{2}{3}\right)^2\right) = 16x + 57 + 4 \)
\( \implies 9\left(y - \frac{2}{3}\right)^2 = 16x + 61 \)
\( \implies \left(y - \frac{2}{3}\right)^2 = \frac{16}{9}x + \frac{61}{9} \)
\( \implies \left(y - \frac{2}{3}\right)^2 = \frac{16}{9}\left(x + \frac{61}{16}\right) \)
यह परवलय के मानक रूप \( Y^2 = 4aX \) के समान है, जहाँ \( Y = y - \frac{2}{3} \) और \( X = x + \frac{61}{16} \)
इस प्रकार के परवलय का अक्ष \( Y = 0 \) होता है। अक्ष, परवलय को दो समान हिस्सों में बांटती है।
\( \implies y - \frac{2}{3} = 0 \)
\( \implies y = \frac{2}{3} \)
या \( 3y - 2 = 0 \)
अतः, परवलय के अक्ष का समीकरण \( 3y - 2 = 0 \) है।
In simple words: We rewrite the parabola's equation by grouping the y-terms and completing the square. This transforms it into a standard form. The axis of symmetry for such a parabola is simply \( Y = 0 \), which gives us the equation \( 3y - 2 = 0 \).

🎯 Exam Tip: For a parabola in the form \( (y-k)^2 = 4a(x-h) \), the axis of symmetry is \( y-k=0 \). Similarly, for \( (x-h)^2 = 4a(y-k) \), the axis is \( x-h=0 \).

Question 31. दीर्घवृत्त \( \frac{x^2 - ax}{a^2} + \frac{y^2 - by}{b^2} = 0 \) के केन्द्र के निर्देशांक लिखिए।
Answer: दिया गया दीर्घवृत्त का समीकरण है: \( \frac{x^2 - ax}{a^2} + \frac{y^2 - by}{b^2} = 0 \)
इसे हम ऐसे लिख सकते हैं:
\( \frac{x^2}{a^2} - \frac{ax}{a^2} + \frac{y^2}{b^2} - \frac{by}{b^2} = 0 \)
\( \implies \frac{x^2}{a^2} - \frac{x}{a} + \frac{y^2}{b^2} - \frac{y}{b} = 0 \)
केन्द्र के निर्देशांक ज्ञात करने के लिए वर्ग पूरा करेंगे।
\( \implies \left(\frac{x^2}{a^2} - \frac{x}{a} + \frac{1}{4}\right) - \frac{1}{4} + \left(\frac{y^2}{b^2} - \frac{y}{b} + \frac{1}{4}\right) - \frac{1}{4} = 0 \)
\( \implies \left(\frac{x}{a} - \frac{1}{2}\right)^2 + \left(\frac{y}{b} - \frac{1}{2}\right)^2 = \frac{1}{4} + \frac{1}{4} \)
\( \implies \left(\frac{x - a/2}{a}\right)^2 + \left(\frac{y - b/2}{b}\right)^2 = \frac{1}{2} \)
\( \implies \frac{(x - a/2)^2}{a^2} + \frac{(y - b/2)^2}{b^2} = \frac{1}{2} \)
इसे मानक दीर्घवृत्त समीकरण \( \frac{(x-h)^2}{A^2} + \frac{(y-k)^2}{B^2} = 1 \) से तुलना करने पर, जहाँ केन्द्र \( (h, k) \) होता है।
यहाँ, \( x - h = x - a/2 \implies h = a/2 \)
और \( y - k = y - b/2 \implies k = b/2 \)
अतः, दीर्घवृत्त के केन्द्र के निर्देशांक \( \left(\frac{a}{2}, \frac{b}{2}\right) \) हैं। यह केन्द्र दीर्घवृत्त के बिलकुल बीच में होता है।
In simple words: We rewrite the given equation by grouping x and y terms and completing the square for each. This helps us transform it into the standard form of an ellipse. From the standard form, we can directly identify the coordinates of the center, which are \( \left(\frac{a}{2}, \frac{b}{2}\right) \).

🎯 Exam Tip: To find the center of any conic section, rearrange the equation and complete the square for the x and y terms to transform it into its standard form \( \frac{(x-h)^2}{A^2} + \frac{(y-k)^2}{B^2} = 1 \). The center will then be \( (h, k) \).

Question 32. रेखा \( x \cos a + y \sin a = p \) के दीर्घवृत्त \( \frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}} =1 \) को स्पर्श करने का प्रतिबन्ध लिखिए।
Answer: दीर्घवृत्त का समीकरण है: \( \frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}} =1 \)
रेखा का समीकरण है: \( x \cos a + y \sin a = p \)
किसी रेखा \( x \cos \theta + y \sin \theta = p \) के दीर्घवृत्त \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) को स्पर्श करने का प्रतिबन्ध होता है:
\( p^2 = a^2 \cos^2 \theta + b^2 \sin^2 \theta \)
इस प्रश्न में \( \theta \) की जगह \( a \) है।
अतः, प्रतिबन्ध है: \( p^2 = a^2 \cos^2 a + b^2 \sin^2 a \)
यह समीकरण बताता है कि रेखा कब दीर्घवृत्त को छूएगी।
In simple words: For a straight line to just touch an ellipse, there's a special relationship between the line's constant term \( p \) and the ellipse's parameters \( a \) and \( b \), along with the angle \( a \). This relationship is \( p^2 = a^2 \cos^2 a + b^2 \sin^2 a \).

🎯 Exam Tip: Remember the standard condition for tangency of a line to an ellipse in both slope form (\( c^2 = a^2m^2 + b^2 \)) and parametric form (\( p^2 = a^2 \cos^2 \theta + b^2 \sin^2 \theta \)).

Question 33. अतिपरवलय का समीकरण लिखिए जिसकी अनुप्रस्थ अक्ष और संयुग्मी अक्ष क्रमशः 4 तथा 5 हैं ।
Answer: अतिपरवलय का अनुप्रस्थ अक्ष (transverse axis) \( 2a = 4 \) दिया गया है।
\( \implies a = 2 \implies a^2 = 4 \)
संयुग्मी अक्ष (conjugate axis) \( 2b = 5 \) दिया गया है।
\( \implies b = \frac{5}{2} \implies b^2 = \frac{25}{4} \)
यदि अनुप्रस्थ अक्ष x-अक्ष के अनुदिश हो, तो अतिपरवलय का मानक समीकरण होता है:
\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
मानों को प्रतिस्थापित करने पर:
\( \implies \frac{x^2}{4} - \frac{y^2}{\frac{25}{4}} = 1 \)
\( \implies \frac{x^2}{4} - \frac{4y^2}{25} = 1 \)
दोनों तरफ 100 से गुणा करने पर (लघुत्तम समापवर्त्य):
\( \implies 25x^2 - 16y^2 = 100 \)
अतः, अतिपरवलय का समीकरण \( 25x^2 - 16y^2 = 100 \) है।
In simple words: We are given the lengths of the transverse and conjugate axes. We use these to find \( a^2 \) and \( b^2 \). Then, we substitute these values into the standard equation of a hyperbola to get the final equation.

🎯 Exam Tip: Remember that for a hyperbola, the equation involves a minus sign between the \( x^2 \) and \( y^2 \) terms, unlike an ellipse. The transverse axis determines the orientation of the hyperbola.

Question 34. अतिपरवलय \( \frac{(x-1)^2}{16} - \frac{(y+2)^2}{9} = 1 \) के केन्द्र के निर्देशांक लिखिए।
Answer: दिया गया अतिपरवलय का समीकरण है: \( \frac{(x-1)^2}{16} - \frac{(y+2)^2}{9} = 1 \)
यह अतिपरवलय के मानक रूप \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) के समान है, जहाँ केन्द्र के निर्देशांक \( (h, k) \) होते हैं।
यहाँ \( (x-1)^2 \) में, \( h = 1 \)
और \( (y+2)^2 \) में, \( k = -2 \)
केन्द्र के निर्देशांक ज्ञात करने के लिए, हम \( x-1=0 \) और \( y+2=0 \) सेट करते हैं।
\( \implies x - 1 = 0 \implies x = 1 \)
\( \implies y + 2 = 0 \implies y = -2 \)
अतः, अतिपरवलय के केन्द्र के निर्देशांक \( (1, -2) \) हैं। केन्द्र अतिपरवलय के दो शाखाओं के बीच का मध्यबिन्दु होता है।
In simple words: The center of a hyperbola in the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) is simply \( (h, k) \). From the given equation, we can directly see that the center is at \( (1, -2) \).

🎯 Exam Tip: For any conic section shifted from the origin, its center (or vertex for parabola) can be directly identified by setting the terms \( (x-h) \) and \( (y-k) \) to zero.