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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

PQ is a double ordinate of the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ such that $$\Delta OPQ$$ is an equilateral triangle, O being the centre of the hyperbola. Then the eccentricity e of the hyperbola satisfies

A
$$1 < e < {2 \over {\sqrt 3 }}$$
B
$$e = {2 \over {\sqrt 3 }}$$
C
$$e = 2\sqrt 3 $$
D
$$e > {2 \over {\sqrt 3 }}$$

পরাবৃত্ত $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ এর একটি দ্বিকোটি হল PQ এবং $$\Delta OPQ$$ একটি সমবাহু ত্রিভুজ (O হল ঐ পরাবৃত্তের কেন্দ্র) । সেক্ষেত্রে পরাবৃত্তের উৎকেন্দ্রতা যে সম্পর্ককে সিদ্ধ করে সেটি হল

A
$$1 < e < {2 \over {\sqrt 3 }}$$
B
$$e = {2 \over {\sqrt 3 }}$$
C
$$e = 2\sqrt 3 $$
D
$$e > {2 \over {\sqrt 3 }}$$
2

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

AB is a variable chord of the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$. If AB subtends a right angle at the origin O, then $${1 \over {O{A^2}}} + {1 \over {O{B^2}}}$$ equals to

A
$${1 \over {{a^2}}} + {1 \over {{b^2}}}$$
B
$${1 \over {{a^2}}} - {1 \over {{b^2}}}$$
C
$${a^2} + {b^2}$$
D
$${a^2} - {b^2}$$

উপবৃত্ত $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ এর AB একটি চলমান জ্যা। যদি AB সরলরেখা O মূলবিন্দুতে সমকোণ উৎপন্ন করে, তবে $${1 \over {O{A^2}}} + {1 \over {O{B^2}}}$$ হবে

A
$${1 \over {{a^2}}} + {1 \over {{b^2}}}$$
B
$${1 \over {{a^2}}} - {1 \over {{b^2}}}$$
C
$${a^2} + {b^2}$$
D
$${a^2} - {b^2}$$
3

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

Let $$P(3\sec \theta ,2\tan \theta )$$ and $$Q(3\sec \phi ,2\tan \phi )$$ be two points on $${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$$ such that $$\theta + \phi = {\pi \over 2},0 < \theta ,\phi < {\pi \over 2}$$. Then the ordinate of the point of intersection of the normals at P and Q is

A
$${{13} \over 2}$$
B
$$ - {{13} \over 2}$$
C
$${5 \over 2}$$
D
$$ - {5 \over 2}$$

মনে কর $${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$$ এর উপরিস্থ দুটি বিন্দু $$P(3\sec \theta ,2\tan \theta )$$ ও $$Q(3\sec \phi ,2\tan \phi )$$ $$\theta + \phi = {\pi \over 2},0 < \theta ,\phi < {\pi \over 2}$$ । সেক্ষেত্রে P ও Q বিন্দুতে অভিলম্বদ্বয়ের ছেদবিন্দুর কোটি হবে

A
$${{13} \over 2}$$
B
$$ - {{13} \over 2}$$
C
$${5 \over 2}$$
D
$$ - {5 \over 2}$$
4

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
The points of intersection of two ellipses $${x^2} + 2{y^2} - 6x - 12y + 20 = 0$$ and $$2{x^2} + {y^2} - 10x - 6y + 15 = 0$$ lie on a circle. The centre of the circle is
A
(8, 3)
B
(8, 1)
C
$$\left( {{8 \over 3},3} \right)$$
D
(3, 8)

Explanation

Given ellipse,

$${x^2} + 2{y^2} - 6x - 12y + 20 = 0$$ and

$$2{x^2} + {y^2} - 10x - 6y + 15 = 0$$

The point of intersection of the given ellipse is

$$({x^2} + 2{y^2} - 6x - 12y + 20) + \lambda (2{x^2} + {y^2} - 10x - 6y + 15) = 0$$

$$(1 + 2\lambda ){x^2} + (2 + \lambda ){y^2} - (6 + 10\lambda )x - (12 + 6\lambda ) + 20 + 5\lambda = 0$$

The points lie on circle

$$\therefore$$ Equation becomes circle

$$\therefore$$ $$1 + 2\lambda = 2 + \lambda \Rightarrow \lambda = 1$$

$$\therefore$$ Equation of circle is

$$3{x^2} + 3{y^2} - 16x - 18y + 35 = 0$$

$$\therefore$$ Centre of circle is $$\left( {{8 \over 3},3} \right)$$
$${x^2} + 2{y^2} - 6x - 12y + 20 = 0$$ ও $$2{x^2} + {y^2} - 10x - 6y + 15 = 0$$ উপবৃত্তদ্বয়ের ছেদবিন্দুগুলি একটি বৃত্তের উপরিস্থ। ঐ বৃত্তের কেন্দ্র হবে
A
(8, 3)
B
(8, 1)
C
$$\left( {{8 \over 3},3} \right)$$
D
(3, 8)

Explanation

উপবৃত্ত দেওয়া হয়েছে,

$${x^2} + 2{y^2} - 6x - 12y + 20 = 0$$ এবং

$$2{x^2} + {y^2} - 10x - 6y + 15 = 0$$

প্রদত্ত উপবৃত্তের ছেদ বিন্দু হল

$$({x^2} + 2{y^2} - 6x - 12y + 20) + \lambda (2{x^2} + {y^2} - 10x - 6y + 15) = 0$$

$$(1 + 2\lambda ){x^2} + (2 + \lambda ){y^2} - (6 + 10\lambda )x - (12 + 6\lambda ) + 20 + 5\lambda = 0$$

বিন্দু গুলি বৃত্তর উপর অবস্থিত

$$\therefore$$ সমীকরণ বৃত্তে পরিণত হয়

$$\therefore$$ $$1 + 2\lambda = 2 + \lambda \Rightarrow \lambda = 1$$

$$\therefore$$ বৃত্তের সমীকরণ হল

$$3{x^2} + 3{y^2} - 16x - 18y + 35 = 0$$

$$\therefore$$ বৃত্তের কেন্দ্র হল $$\left( {{8 \over 3},3} \right)$$

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