JEE Main 2015 · 10 Apr · Shift 1 · Mathematics Q81

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MathematicsSingle correct+4 / -1Question 81 of 90

If the tangent to the conic, $y-6=x^2$ at (2, 10) touches the circle, $x^2+y^2+8x-2y=k$ (for some fixed k) at a point (α, β) ; then (α, β) is :

\left(-\dfrac6{17},\dfrac{10}{17}\right)

\left(-\dfrac8{17},\dfrac2{17}\right)

\left(-\dfrac4{17},\dfrac1{17}\right)

\left(-\dfrac7{17},\dfrac6{17}\right)

Answer and worked solution

Official answer: B

The tangent at (2, 10) has slope $2x=4$ : $y=4x+2$ . The circle's centre is $(-4,1)$ , and the point of contact is the foot of the perpendicular from the centre to $4x-y+2=0$ :
$t=-\frac{4(-4)-1+2}{17}=\frac{15}{17},\qquad(\alpha,\beta)=\left(-4+\frac{60}{17},\,1-\frac{15}{17}\right)=\left(-\frac8{17},\frac2{17}\right).$

Topics:Tangents and normals Parabola

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JEE Main 2015 · 10 Apr · Shift 1, Mathematics question 81