Ellipse 2 Question 5

5. If the tangents on the ellipse 4x2+y2=8 at the points (1,2) and (a,b) are perpendicular to each other, then a2 is equal to

(a) 12817

(b) 6417

(c) 417

(d) 217

(2019 Main, 8 April I)

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Solution:

  1. Equation of given ellipse is

4x2+y2=8 x22+y28=1x2(2)2+y2(22)2=1

Now, equation of tangent at point (1,2) is

2x+y=4

[ equation of tangent to the ellipse x2a2+y2b2=1 at (x1,y1)

is xx1a2+yy1b2=1]

and equation of another tangent at point (a,b) is

4ax+by=8

Since, lines (ii) and (iii) are perpendicular to each other.

21×4ab=1

[if lines a1x+b1y+c1=0 and a2x+b2y+c2=0

are perpendicular, then a1b1a2b2=1 ]

b=8a

Also, the point (a,b) lies on the ellipse (i), so

4a2+b2=8

4a2+64a2=868a2=8a2=868a2=217

[from Eq.(iv)]



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