If the line αx+4y= 7, where ∈R, touches the ellipse 3x^2+4y^2=1 at the point P in the first quadrant, then one of the focal distances of P is :

Question

Accepted Answer

Correct answer: A. (a) Ellipse 3x^2+4y^2=1 :a^2= 13 ,b^2= 14 ,e= 12. Tangent at (x_0 ,y_0 ) :3x_0x+4y_0y=1. Comparing with αx+4y= 7 : 4y_0= 4 7 ⇒y_0= 1 7 and 3x_0= 7. Substituting in ellipse equation: ^221+ 47=1 ⇒ ^2=9 ⇒ =3. P= ( 1 7 , 1 7 ). Focal distances =a ex_0= 1 3 12 7. One focal distance is 1 3+ 12 7.

$If the line αx + 4 y = 7, where α \in R, touches the ellipse 3 x^{2} + 4 y^{2} = 1 at the point P in the first quadrant, then one of the focal distances of P is :$

Solution

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