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 :
Solutions · 1
Official solution
Given, Equation of ellipse 4x2
+ y2 = 8
We know equation of tangent at any point (x1, y1) is
4xx1
+ yy1 = 8
\[ \therefore \] Equation of tangent at point (1, 2) is
4x + 2y = 8
\[ \Rightarrow \] 2x + y = 4
\[ \therefore \] Slope of this tangent = -2
Equation of tangent at point (a, b) is
4ax + by = 8
Slope of this tangent = \[ - {{4a} \over b}\]
As tangent at (1, 2) and (a, b) are perpendicular
\[ \therefore \] \[\left( { - 2} \right)\left( { - {{4a} \over b}} \right) = - 1\]
\[ \Rightarrow \] b = -8a
As point (a, b) lies on the ellipse
\[ \therefore \] 4a2+b2=8
\[ \Rightarrow \] 4a2+64a2=8
\[ \Rightarrow \] \[{a^2} = {8 \over {68}}\] = \[{2 \over {17}}\]