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1

WB JEE 2022

MCQ (More than One Correct Answer)
English
Bengali

Chords of an ellipse are drawn through the positive end of the minor axis. Their midpoint lies on

A
a circle
B
a parabola
C
an ellipse
D
a hyperbola

উপবৃত্তের উপাক্ষর ধনাত্মক অংশের প্রান্তবিন্দু থেকে উপবৃত্তের জ্যাগুলি অঙ্কিত হল। জ্যাগুলির মধ্যবিন্দুসমূহের সঞ্চারপথ হল

A
একটি বৃত্ত
B
একটি অধিবৃত্ত
C
একটি উপবৃত্ত
D
একটি পরাবৃত্ত
2

WB JEE 2020

MCQ (More than One Correct Answer)
English
Bengali
Consider a tangent to the ellipse $${{{x^2}} \over 2} + {{{y^2}} \over 1} = 1$$ at any point. The locus of the mid-point of the portion intercepted between the axes is
A
$${{{x^2}} \over 2} + {{{y^2}} \over 4} = 1$$
B
$${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$$
C
$${1 \over {3{x^2}}} + {1 \over {4{y^2}}} = 1$$
D
$${1 \over {2{x^2}}} + {1 \over {4{y^2}}} = 1$$

Explanation

Equation of tangent to the ellipse $${{{x^2}} \over 2} + {{{y^2}} \over 1} = 1$$ at point (x1, y1) is

$${{x{x_{^2}}} \over 2} + {{y{y_{^2}}} \over 1} = 1$$ ....(i)

Let mid-point of intercept be P(h, k).

$$ \therefore $$ $$h = {1 \over {{x_1}}}$$

$$ \Rightarrow {x_1} = {1 \over h}$$

and $$k = {1 \over {2{y_1}}}$$

$$ \Rightarrow {y_1} = {1 \over {2k}}$$

$$ \therefore $$ Required locus of the mid-point

$$ = {1 \over {2{x^2}}} + {1 \over {4{y^2}}} = 1$$

উপবৃত্ত $${{{x^2}} \over 2} + {{{y^2}} \over 1} = 1$$-এর উপরিস্থ যেকোনো বিন্দুতে স্পর্শক বিবেচনা করো। অক্ষদ্বয়ের মধ্যে ওই স্পর্শকের ছেদিতাংশের মদ্ধবিন্দুর সঞ্চারপথ হবে

A
$${{{x^2}} \over 2} + {{{y^2}} \over 4} = 1$$
B
$${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$$
C
$${1 \over {3{x^2}}} + {1 \over {4{y^2}}} = 1$$
D
$${1 \over {2{x^2}}} + {1 \over {4{y^2}}} = 1$$

Explanation

উপবৃত্তের উপর যে কোন বিন্দু $$(\sqrt 2 \cos \varphi ,\sin \varphi )$$

$$\therefore$$ স্পর্শকের সমীকরণ $${{x\,.\,\sqrt 2 \cos \varphi } \over 2} + {{y\,.\,\sin \varphi } \over 1} = 1$$

বা, $${x \over {{{\sqrt 2 } \over {\cos \varphi }}}} + {y \over {{1 \over {\sin \varphi }}}} = 1$$

মধ্যবিন্দু $$P(h,k)$$ হলে

$$h = {1 \over {\sqrt 2 \cos \varphi }}$$ এবং $$k = {1 \over {2\sin \varphi }}$$

$$\because$$ $${\sin ^2}\varphi + {\cos ^2}\varphi = 1$$

$$\therefore$$ $${1 \over {2{h^2}}} + {1 \over {4{k^2}}} = 1$$

$$\therefore$$ সঞ্চার পথের সমীকরণ $${1 \over {2{x^2}}} + {1 \over {4{y^2}}} = 1$$

3

WB JEE 2019

MCQ (More than One Correct Answer)
English
Bengali
Equation of a tangent to the hyperbola 5x2 $$-$$ y2 = 5 and which passes through an external point (2, 8) is
A
3x $$-$$ y + 2 = 0
B
3x + y $$-$$ 14 = 0
C
23x $$-$$ 3y $$-$$ 22 = 0
D
3x $$-$$ 23y + 178 = 0

Explanation

Given equation of hyperbola can be write as

$${{{x^2}} \over 1} - {{{y^2}} \over {{{(\sqrt 5 )}^2}}} = 1$$

here, a = 1 and b = $${\sqrt 5 }$$

Since, $$y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} $$

$$y = mx \pm \sqrt {{m^2} - 5} $$

$$ \Rightarrow {(8 - 2m)^2} = {(\sqrt {{m^2} - 5} )^2}$$

$$ \Rightarrow {(8 - 2m)^2} = {m^2} - 5$$

$$ \Rightarrow 64 + 4{m^2} - 32m = {m^2} - 5$$

$$ \Rightarrow 3{m^2} - 32m + 69 = 0$$

$$ \Rightarrow m = {{22} \over 3},3$$ [by solving equation]

When $$m = {{22} \over 3}$$, then equation of tangent

$$(y - 8) = {{22} \over 3}(x - 2)$$

$$ \Rightarrow 3y - 24 = 23x - 46$$

$$ \Rightarrow 23x - 3y - 46 + 24 = 0$$

$$ \Rightarrow 23x - 3y - 22 = 0$$

When m = 3, then equation of tangent

$$(y - 8) = 3(x - 2)$$

$$ \Rightarrow y - 8 = 3x - 6$$

$$ \Rightarrow 3x - y - 6 + 8 = 0$$

$$ \Rightarrow 3x - y + 2 = 0$$

Hence, option (a) and (c) are correct.

$$5{x^2} - {y^2} = 5$$ পরাবৃত্তের একটি স্পর্শক বহিঃস্থ বিন্দু $$(2,8)$$ দিয়ে যায়। ওই স্পর্শকের সমীকরণ হবে -

A
$$3x - y + 2 = 0$$
B
$$3x + y - 14 = 0$$
C
$$23x - 3y - 22 = 0$$
D
$$3x - 23y + 178 = 0$$

Explanation

স্পর্শকের সমীকরণ $$y = mx \pm \sqrt {{m^2} - 5} $$ ইহা $$(2,8)$$ বিন্দুগামী।

$$\therefore$$ $$8 = 2m \pm \sqrt {{m^2} - 5} $$

বা, $${(8 - 2m)^2} = {m^2} - 5$$

বা, $$64 - 32m + 4{m^2} = {m^2} - 5$$

বা, $$3{m^2} - 32m + 69 = 0$$

বা, $$3{m^2} - 23m - 9m + 69 = 0$$

বা, $$m(3m - 23) - 3(3m - 23) = 0$$

$$\therefore$$ $$m = 3,{{23} \over 3}$$

$$\therefore$$ স্পর্শকের সমীকরণ হবে, $$y = 3x \pm \sqrt {9 - 5} $$ অথবা $$y = {{23} \over 3}x \pm \sqrt {{{\left( {{{23} \over 3}} \right)}^2} - 5} $$

$$\therefore$$ বা, $$y = 3x \pm 2$$ অথবা $$3y = 23x \pm 22$$

4

WB JEE 2018

MCQ (More than One Correct Answer)
English
Bengali
A hyperbola, having the transverse axis of length 2sin$$\theta$$ is confocal wit6h the ellipse 3x2 + 4y2 = 12. Its equation is
A
x2sin2$$\theta$$ $$-$$ y2cos2$$\theta$$ = 1
B
x2cosec2$$\theta$$ $$-$$ y2sec2$$\theta$$ = 1
C
(x2 + y2)sin2$$\theta$$ = 1 + y2
D
x2cosec2$$\theta$$ = x2 + y2 + sin2$$\theta$$

Explanation

Given, $$2{a_1} = 2\sin \theta $$

$$ \Rightarrow {a_1} = \sin \theta $$

and $$3{x^2} + 4{y^2} = 12$$

$$ \Rightarrow {{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$$

Here, $${a^2} = 4$$ and $${b^2} = 3$$

$$ \therefore $$ $${b^2} = {a^2}(1 - {e^2})$$

$$ \Rightarrow 3 = 4(1 + {e^2})$$

$$ \Rightarrow {e^2} = 1 - {3 \over 4} = {1 \over 4}$$

$$ \Rightarrow e = {1 \over 2}$$

Focus, $$F(ae,0) = F\left( {2 \times {1 \over 2},0} \right)$$

$$ = F(1,0)$$

For hyperbola foci are same

$${a_1}{e_1} = ae = 1$$

$$ \therefore $$ $$(\sin \theta ){e_1} = 1$$

$$ \Rightarrow {e_1} = \cos ec\theta $$

and $$b_1^2 = a_1^2(e_1^2 - 1) = a_1^2e_1^2 - a_1^2$$

$$ \Rightarrow b_1^2 = 1 - {\sin ^2}\theta = {\cos ^2}\theta $$

$${{{x^2}} \over {a_1^2}} - {{{y^2}} \over {b_1^2}} = 1$$

$$ \Rightarrow {{{x^2}} \over {{{\sin }^2}\theta }} - {{{y^2}} \over {{{\cos }^2}\theta }} = 1$$

$$ \Rightarrow {x^2}\cos e{c^2}\theta - {y^2}{\sec ^2}\theta = 1$$

একটি পরাবৃত্তের অনুপ্রস্থ অক্ষের দৈর্ঘ্য 2sin$$\theta$$ । পরাবৃত্তটি $$3{x^2} + 4{y^2} = 12$$ উপবৃত্তের সমনাভি। ইহার সমীকরণ হল -

A
$${x^2}{\sin ^2}\theta - {y^2}{\cos ^2}\theta = 1$$
B
$${x^2}\cos e{c^2}\theta - {y^2}{\sec ^2}\theta = 1$$
C
$$({x^2} + {y^2}){\sin ^2}\theta = 1 + {y^2}$$
D
$${x^2}\cos e{c^2}\theta = {x^2} + {y^2} + {\sin ^2}\theta $$

Explanation

$${{{x^2}} \over 4} + {{{y^3}} \over 3} = 1$$ উপবৃত্তের উৎকেন্দ্রতা

$$\sqrt {1 - {3 \over 4}} = {1 \over 2}$$

নাভি $$( \pm 1,0)$$

পরাবৃত্তের সমীকরণ $${{{x^2}} \over {{{\sin }^2}\theta }} - {{{y^2}} \over {{b^2}}} = 1$$

$$e = \sqrt {1 + {{{b^2}} \over {{{\sin }^2}\theta }}} $$

নাভি $$ = \left( { \pm \sin \theta \sqrt {1 + {{{b^2}} \over {{{\sin }^2}\theta }}} ,0} \right) = \left( { \pm \sqrt {{{\sin }^2}\theta + {b^2}} ,0} \right)$$

$$\therefore$$ $$\sqrt {{{\sin }^2}\theta + {b^2}} = 1 \Rightarrow {b^2} = {\cos ^2}\theta $$

$$\therefore$$ পরাবৃত্তের সমীকরণ $${{{x^2}} \over {{{\sin }^2}\theta }} - {{{y^2}} \over {{{\cos }^2}\theta }} = 1$$

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