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

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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

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