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Question of 148

Q.Find the equations of the tangents to the hyperbola : 3x2−4y2=123x^2 - 4y^2 = 12 which are :

(i) Parallel and
(ii) Perpendicular to the line y=x−7y = x - 7.
Telangana TsbieTelangana Board of Intermediate Education 2020Subjective· 4mImportance★★★★★
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Put the hyperbola in standard form to get a2,b2a^2,b^2, then use the tangent-in-terms-of-slope formula y=mx±a2m2−b2y=mx\pm\sqrt{a^2m^2-b^2} with the required slope mm.

Hyperbola: 3x2−4y2=12⇒x24−y23=13x^2-4y^2=12 \Rightarrow \dfrac{x^2}{4}-\dfrac{y^2}{3}=1, so a2=4, b2=3a^2=4,\ b^2=3.

Line y=x−7y=x-7 has slope 11.

(i) Parallel (m=1m=1):

y=mx±a2m2−b2=x±4(1)−3=x±1y=mx\pm\sqrt{a^2m^2-b^2}=x\pm\sqrt{4(1)-3}=x\pm1

Tangents: y=x+1y=x+1 and y=x−1y=x-1.

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