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Q.Find the value of kk, if 4x+y+k=04x + y + k = 0 is a tangent to the ellipse x2+3y2=3x^2 + 3y^2 = 3.

Telangana TsbieTelangana Board of Intermediate Education 2025Subjective· 4mImportance★★★★★
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For x23+y2=1\tfrac{x^2}{3}+y^2=1, a2=3,b2=1a^2=3,b^2=1; with m=−4,c=−km=-4,c=-k, c2=a2m2+b2⇒k2=49c^2=a^2m^2+b^2\Rightarrow k^2=49.

Write the ellipse in standard form: x2+3y2=3⇒x23+y21=1x^2+3y^2=3\Rightarrow \dfrac{x^2}{3}+\dfrac{y^2}{1}=1, so a2=3, b2=1a^2=3,\ b^2=1.

The line 4x+y+k=04x+y+k=0 is y=−4x−ky=-4x-k, i.e. slope m=−4m=-4, intercept c=−kc=-k.

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