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# Engineering

The tangent to the parabola y^{2}=4xat the point where it intersects the circle x^{2}+y^{2}=5 in the first quadrant, passes through the point :

  • Option 1)

    \left ( -\frac{1}{3},\frac{4}{3} \right )

  • Option 2)

    \left ( \frac{1}{4},\frac{3}{4} \right )

  • Option 3)

    \left ( \frac{3}{4},\frac{7}{4} \right )

  • Option 4)

    \left ( -\frac{1}{4},\frac{1}{2} \right )

 

Answers (1)
S solutionqc

Point of intersection of y^{2}=4x and x^{2}+y^{2}=5 is (1,2) and (1,-2)

Equation of tangent at (1, 2)

 2y=2(x+1)\Rightarrow y=x+1

y-x=1    point \left ( \frac{3}{4} , \frac{7}{4}\right ) satisfies .

 


Option 1)

\left ( -\frac{1}{3},\frac{4}{3} \right )

Option 2)

\left ( \frac{1}{4},\frac{3}{4} \right )

Option 3)

\left ( \frac{3}{4},\frac{7}{4} \right )

Option 4)

\left ( -\frac{1}{4},\frac{1}{2} \right )

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