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Mathematics Question on Conic sections

The locus of the point of intersection of the tangents drawn at the ends of a focal chord of the parabola x2=−8yx^2=-8y is _______

A

y = 2

B

y = -2

C

x = 2

D

x = -2

Answer

y = 2

Explanation

Solution

Given that, parabola x2=−8yx^{2}=-8 y

Here, on comparing with x2=4ayx^{2}=4 a y
⇒4a=−8⇒a=−2\Rightarrow 4 a=-8 \Rightarrow a=-2
Let the parametric coordinate of parabola
x2=−8yx^{2}=-8 y is, P→(4t,−2t2)P \rightarrow\left(4 t,-2 t^{2}\right)
and the other coordinate of latusrectum is
P′→(−4t,−2t2)P' \rightarrow\left(\frac{-4}{t}, \frac{-2}{t^{2}}\right)
Now, the equation tangent of parabola x2=−8yx^{2}=-8 y
x⋅x1=−4(y+y1)x \cdot x_{1}=-4\left(y+y_{1}\right)...(i)
At Px(4t)=−4(y−2t2)P x(4 t)=-4\left(y-2 t^{2}\right)
xt=−y+2t2x t=-y+2 t^{2}
xt+y=2t2x t+ y=2 t^{2}...(ii)
⇒x(−4t)=−4(y−2t2)\Rightarrow x\left(\frac{-4}{t}\right)=-4\left(y-\frac{2}{t^{2}}\right)
⇒xt=y−2t2\Rightarrow \frac{x}{t}=y-\frac{2}{t^{2}}
⇒xt=yt2−2\Rightarrow x t=y t^{2}-2
⇒xt−yt2=−2\Rightarrow x t-y t^{2}=-2...(iii)
On solving Eqs. (i) and (ii)
xt3+yt2=2t4x t^{3}+y t^{2} =2 t^{4}
xt−yt2=−2x t-y t^{2} =-2

xt(1+t2)=−2(1−t4)x t\left(1+t^{2}\right) =-2\left(1-t^{4}\right)
xt(1+t2)=−2(1+t2)(1−t2)x t\left(1+t^{2}\right) =-2\left(1+t^{2}\right)\left(1-t^{2}\right)
tx=−2(1−t2)t x =-2\left(1-t^{2}\right)...(iv)
From E (ii)
−2(1−t2)+y=2t2-2\left(1-t^{2}\right)+y =2 t^{2}
⇒−2+2t2+y=2t2\Rightarrow -2+2 t^{2}+y =2 t^{2}
⇒y=2\Rightarrow y =2
Hence, the intersection point of both tangent lying on QQ.
ie, y=2.y=2 . Which is the required locus.