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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

Two circles $${S_1} = p{x^2} + p{y^2} + 2g'x + 2f'y + d = 0$$ and $${S_2} = {x^2} + {y^2} + 2gx + 2fy + d' = 0$$ have a common chord PQ. The equation of PQ is

A
$${S_1} - {S_2} = 0$$
B
$${S_1} + {S_2} = 0$$
C
$${S_1} - p{S_2} = 0$$
D
$${S_1} + p{S_2} = 0$$

দুটি বৃত্ত $${S_1} = p{x^2} + p{y^2} + 2g'x + 2f'y + d = 0$$ ও $${S_2} = {x^2} + {y^2} + 2gx + 2fy + d' = 0$$ এর একটি সাধারণ জ্যা PQ আছে। তবে PQ এর সমীকরণ হবে

A
$${S_1} - {S_2} = 0$$
B
$${S_1} + {S_2} = 0$$
C
$${S_1} - p{S_2} = 0$$
D
$${S_1} + p{S_2} = 0$$
2

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

The side AB of $$\Delta$$ABC is fixed and is of length 2a unit. The vertex moves in the plane such that the vertical angle is always constant and is $$\alpha$$. Let x-axis be along AB and the origin be at A. Then the locus of the vertex is

A
$${x^2} + {y^2} + 2ax\sin \alpha + {a^2}\cos \alpha = 0$$
B
$${x^2} + {y^2} - 2ax - 2ay\cot \alpha = 0$$
C
$${x^2} + {y^2} - 2ax\cos \alpha - {a^2} = 0$$
D
$${x^2} + {y^2} - ax\sin \alpha - ay\cos \alpha = 0$$

$$\Delta$$ABC ত্রিভুজের AB বাহু অনড় ও 2a একক দৈর্ঘ্য সম্পন্ন। শীর্ষ কৌণিক বিন্দুটি ঐ তলে এরূপ ভাবে চলমান যে শীর্ষকোণটি সর্বদাই ধ্রুবক $$\alpha$$ হবে। মনে কর ভুমি রেখা AB বরাবর x-অক্ষ রয়েছে ও মূলবিন্দুটি A-তে রয়েছে। সেক্ষেত্রে শীর্ষবিন্দুর সঞ্চারপথ হবে

A
$${x^2} + {y^2} + 2ax\sin \alpha + {a^2}\cos \alpha = 0$$
B
$${x^2} + {y^2} - 2ax - 2ay\cot \alpha = 0$$
C
$${x^2} + {y^2} - 2ax\cos \alpha - {a^2} = 0$$
D
$${x^2} + {y^2} - ax\sin \alpha - ay\cos \alpha = 0$$
3

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

If the equation of one tangent to the circle with centre at (2, $$-$$1) from the origin is 3x + y = 0, then the equation of the other tangent through the origin is

A
$$3x - y = 0$$
B
$$x + 3y = 0$$
C
$$x - 3y = 0$$
D
$$x + 2y = 0$$

একটি বৃত্তের কেন্দ্র (2, $$-$$1) দেওয়া আছে। ঐ বৃত্তের মূলবিন্দু থেকে অঙ্কিত একটি স্পর্শকের সমীকরণ হল $$3x + y = 0$$ । সেক্ষেত্রে মূলবিন্দু থেকে অঙ্কিত অপর স্পর্শকের সমীকরণ হবে

A
$$3x - y = 0$$
B
$$x + 3y = 0$$
C
$$x - 3y = 0$$
D
$$x + 2y = 0$$
4

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

A curve passes through the point (3, 2) for which the segment of the tangent line contained between the co-ordinate axes is bisected at the point of contact. The equation of the curve is

A
$$y = {x^2} - 7$$
B
$$x = {{{y^2}} \over 2} + 2$$
C
$$xy = 6$$
D
$${x^2} + {y^2} - 5x + 7y + 11 = 0$$

Explanation

According to the question, p(x, y) is the midpoint of line AB.

$$\therefore$$ $${{\alpha + 0} \over 2} = x \Rightarrow \alpha = 2x$$

$${{\beta + 0} \over 2} = y \Rightarrow \beta = 2y$$

$$\therefore$$ Point A = (2x, 0)

and Point B = (0, 2y)

Slope of the tangent,

$${{dy} \over {dx}} = {{2y - 0} \over {0 - 2x}}$$

$$ \Rightarrow {{dy} \over {dx}} = - {y \over x}$$

$$ \Rightarrow x\,dy + y\,dx = 0$$

$$ \Rightarrow d(xy) = 0$$

$$ \Rightarrow xy = c$$

This curve goes through point (3, 2). So this point satisfy the equation.

$$\therefore$$ 3 . 2 = c

$$\Rightarrow$$ c = 6

$$\therefore$$ Equation of the curve,

xy = 6

একটি বক্ররেখা (3, 2) বিন্দুগামী, বক্ররেখাটির একটি বিন্দুতে অঙ্কিত স্পর্শকের অক্ষদ্বয়ের মধ্যেকার ছেদিতাংশ ঐ স্পর্শবিন্দুতে সমদ্বিখন্ডিত হয়। বক্ররেখাটির সমীকরণ হবে

A
$$y = {x^2} - 7$$
B
$$x = {{{y^2}} \over 2} + 2$$
C
$$xy = 6$$
D
$${x^2} + {y^2} - 5x + 7y + 11 = 0$$

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