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 :

Question

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) (64/17) (B) (2/17) (C) (128/17) (D) (4/17)

Tardigrade Admin · Accepted Answer

4a2 +b2 = 8 .......(1) also (dy/dx) Bigg)(1,2) = - ( 4x/y) =-2 ⇒ - (4a/b) = (1/2) b=-8a ⇒ b2 =64a2 68a2=8 a2 = (2/17)

Q. If the tangents on the ellipse $4 x^{2} + y^{2} = 8$ at the points $(1, 2)$ and $(a, b)$ are perpendicular to each other, then $a^{2}$ is equal to :

$\frac{64}{17}$

$\frac{2}{17}$

$\frac{128}{17}$

$\frac{4}{17}$

$4 a^{2} + b^{2} = 8$ .......(1)
also $\frac{d y}{d x})_{(1, 2)} = - \frac{4 x}{y} = - 2$
$\Rightarrow - \frac{4 a}{b} = \frac{1}{2} b = - 8 a$
$\Rightarrow b^{2} = 64 a^{2} 68 a^{2} = 8$
$a^{2} = \frac{2}{17}$

Q. 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 :

1764​

172​

17128​

174​