The number of values of c such that the strai | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The line $y = 4x + c$ touches the ellipse $\frac{x^2}{4} + y^2 = 1$.Substituting $y = 4x + c$ into the equation of the ellipse:$\frac{x^2}{4} + (4

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The line $y = 4x + c$ touches the ellipse $\frac{x^2}{4} + y^2 = 1$.Substituting $y = 4x + c$ into the equation of the ellipse:$\frac{x^2}{4} + (4x + c)^2 = 1$$x^2 + 4(16x^2 + 8cx + c^2) = 4$$x^2 + 64x^2 + 32cx + 4c^2 - 4 = 0$$65x^2 + 32cx + (4c^2 - 4) = 0$For the line to touch the curve,the discriminant $\Delta$ must be $0$:$\Delta = (32c)^2 - 4(65)(4c^2 - 4) = 0$$1024c^2 - 16(65)(c^2 - 1) = 0$Divide by $16$:$64c^2 - 65(c^2 - 1) = 0$$64c^2 - 65c^2 + 65 = 0$$-c^2 + 65 = 0$$c^2 = 65$$c = \pm \sqrt{65}$Thus,there are $2$ possible values for $c$.

The number of values of $c$ such that the straight line $y = 4x + c$ touches the curve $\frac{x^2}{4} + y^2 = 1$ is

Similar Questions

The ellipse $4x^2 + 9y^2 = 1$ intersects the positive $Y$-axis at point $A$. If $S$ and $S'$ are its foci, then the area (in sq. units) of $\Delta SAS'$ is

If the line $y = 2x + c$ is a tangent to the ellipse $\frac{x^2}{8} + \frac{y^2}{4} = 1$,then $c = $

If the length of the minor axis of an ellipse is equal to one-fourth of the distance between the foci,then the eccentricity of the ellipse is:

If the origin is the centre,the $X$-axis is the major axis,and $\sqrt{\frac{2}{5}}$ is the eccentricity of an ellipse which passes through $(-3, 1)$,then the equation of that ellipse is:

If the curves $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{25}+\frac{y^2}{16}=1$ cut each other orthogonally, then $a^2-b^2$ equals to

Vedclass Test Series

Exam Paper Generator

Online Exam Module