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Questions  

 A common tangent to the conics x2=6y and 2x24y2=9 is 

a
x+y=1
b
x−y=1
c
x+y=92
d
x−y=32

detailed solution

Correct option is D

Let y=(mx+c) is tangent to x2=6ysolve the above equations ⇒x2=6(mx+c) So, x2−6mx−6c=0 Put D=b2−4ac=0⇒c=−32m2∴ we get y=mx−32m2…….(1) And given hyperbola equation is 2x2−4y2=9⇒x292−y294=1….(2) Since, equation (1) is a tangent of equation (2) then c2=a2m2−b2⇒94m4=92m2−94⇒m4=2m2−1⇒m4−2m2+1=0⇒m2−12=0⇒m=±1 for m=1 , equation of tangent is x−y=32Therefore, the correct answer is (D).

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Similar Questions

Statement-1: The locus of the point of intersection of the tangents that are at right angles to the hyperbola x236-y216=1 is the circle x2+y2=52

Statement-2: Perpendicular tangents to the hyperbola x2a2-y2b2=1 interest on the director circle  x2+y2=a2-b2a2>b2 of the hyperbola.


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