Engineering
Mathematics
Standard Parabola
Tangent to Ellipse
Question

A normal with slope 16 is drawn from the point (0, – α) to the parabola x2 = – 4ay, where a > 0. Let L be the line passing through (0, – α) and parallel to the directrix of the parabola. Suppose that L intersects the parabola at two points A and B. Let r denote the length of the latus rectum and s denote the square of the length of the line segment AB. If r : s = 1 : 16, then the value of 24a is ____.

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Solution

dydx=x2adydxN=t

Slope of normal t=1t=16t=6

Now, at2+α2at=1t

⇒  – at2 + α = 2a

⇒ – 6a + α = 2a ⇒ α = 8a

For A and B

x2 = – 4a(– 8a)

x2=32a2x=±42a

A(42a,8a),B(42a,8a)

AB2=(82a)2=128a2=s

∴ Length of LR = r = 4a

rs=4a128a2=116

32a=16a=12

∴ 24a = 12 Ans.