JEE Main 2022 · 30 Jul · Shift 2 · Mathematics Q86

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MathematicsInteger+4 / -1Question 86 of 90

Let A(0, a + 2), B(0, a), C(−2, 0) and D(2, 0) be four points and the lines AD and BC intersect at P(x, y). If the locus of P is the curve f(x, y) = 0 and the tangent at the point (4, γ) on this curve is $\dfrac x\alpha+\dfrac y\beta=1$ , then α (β − γ) is equal to _______.

Your answer

Answer and worked solution

Official answer: 4

Locus. AD: $\frac x2+\frac y{a+2}=1$ ; BC: $-\frac x2+\frac ya=1$ . Eliminating a: $x^2+2xy=4$ .

At x = 4: $\gamma=-\frac32$ . Slope $y'=-\frac{x+y}x=-\frac58$ , tangent $\frac{5x}8+y=1$ : $\alpha=\frac85$ , $\beta=1$ .

Value $\frac85\left(1+\frac32\right)=4$ .

Answer: 4

Topics:Locus Tangents and normals

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JEE Main 2022 · 30 Jul · Shift 2, Mathematics question 86