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

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

Question

(c) The line $y = 4x + c$ touches the ellipse $\frac{{{x^2}}}{4} + {y^2} = 1$

$	herefore \frac{{{x^2}}}{4} + {(4x + c)^2} - 1 = 0$ ==> $65{x^2}

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(c) The line $y = 4x + c$ touches the ellipse $\frac{{{x^2}}}{4} + {y^2} = 1$

$	herefore \frac{{{x^2}}}{4} + {(4x + c)^2} - 1 = 0$ ==> $65{x^2} + 32cx + 4{c^2} - 4 = 0$

has equal roots. Therefore $\Delta = 0$

$	herefore 64{c^2} - 4.65.4({c^2} - 1) = 0$ ==>$4{c^2} - 65{c^2} + 65 = 0$

==> ${c^2} = \frac{{65}}{{61}}$==> $c = \pm \sqrt {\frac{{65}}{{61}}} $ .

Hence, there are two values of $c.$

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 eccentricity of the ellipse $25{x^2} + 16{y^2} = 100$, is

The equations of the common tangents to the ellipse, $ x^2 + 4y^2 = 8 $ $\&$ the parabola $y^2 = 4x$ can be

Eccentricity of the ellipse $9{x^2} + 25{y^2} = 225$ is

The foci of the ellipse $25{(x + 1)^2} + 9{(y + 2)^2} = 225$ are at

The foci of $16{x^2} + 25{y^2} = 400$ are