JEE Main 2022MathematicsHyperbolaMediumNumerical

JEE Main 2022Hyperbola Question with Solution

JEE Main 2022 (25 Jul Shift 1)

Question

Let the equation of two diameters of a circle x2+y2-2x+2fy+1=0 be 2px-y=1 and 2x+py=4p. Then the slope m0, of the tangent to the hyperbola 3x2-y2=3 passing through the centre of the circle is equal to _____.

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Show full solutionCorrect answer: 2
Correct answer
2

Step-by-step explanation

Given, 

Equation of circle x2+y2-2x+2fy+1=0 and given diametric lines will pass through 1,-f which is the centre of the circle, so 2p+f-1=0      ...1 and 2-pf-4p=0       ...2

Now solving equation 1 & 2 we get,  f=0 or -3

Now given, Hyperbola 3x2-y2=3 or x2-y23=1

We know that tangent to hyperbola is given by y=mx±m2-3

It passes 1,0 0=m±m2-3

m tends

Also given the tangent passes through centre 1,3

So, 3=m±m2-3 3-m2=m2-3

m=2

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About this question

This is a previous-year question from JEE Main 2022, covering the Hyperbola chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.