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KEAM 2017 Math Paper

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Question : 41 of 120

Marks: +1, -0

f the straight line y = 4x + c touches the ellipse

\frac{x^2}{4}+y^2=1

then c is equal to

0
± $\sqrt{65}$
± $\sqrt{62}$
± $\sqrt{2}$
±13

Solution:

We have

y=4x+c

....(i)

and

\;\frac{x^{2}}{4}+y^{2}=1

.....(ii)

Put value of

y

from Eqs. (i) into (ii), we get

\;\frac{x^{2}}{4}+(4x+c)^{2}=1

\Rightarrow\;\; x^{2}+4(4x+c)^{2}=4

\Rightarrow\;\; x^{2}+4(16x^{2}+8cx+c^{2})=4

\Rightarrow\;\; x^{2}+64x^{2}+32cx+4c^{2}=4

\Rightarrow\;\; 65x^{2}+32cx+4(c^{2}-1)=0

since, given line is a tangent to the ellipse.

\therefore

Discriminant

=0

\Rightarrow\;\; (32c)^{2}-4 \times 65 \times 4(c^{2}-1)=0

\Rightarrow\;\; 1024c^{2}-1040(c^{2}-1)=0

\Rightarrow\;\; 16c^{2}=1040

\Rightarrow\;\; c^{2}=65

\Rightarrow

c=\pm \sqrt{65}

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