JEE Main & Advanced Mathematics Conic Sections Question Bank Parabola

  • question_answer
    If The tangent to the parabola \[{{y}^{2}}=ax\] makes an angle of 45o with x-axis, then the point of contact is   [RPET 1985, 90, 2003]

    A)            \[\left( \frac{a}{2},\ \frac{a}{2} \right)\]                                    

    B)            \[\left( \frac{a}{4},\ \frac{a}{4} \right)\]

    C)            \[\left( \frac{a}{2},\ \frac{a}{4} \right)\]                                    

    D)            \[\left( \frac{a}{4},\ \frac{a}{2} \right)\]

    Correct Answer: D

    Solution :

               Parabola is \[{{y}^{2}}=ax\]i.e., \[{{y}^{2}}=4\left( \frac{a}{4} \right)\,x\]    .....(i)                    \[\because \] Let point of contact is \[({{x}_{1}},\,{{y}_{1}})\]                    \ Equation of tangent is \[y-{{y}_{1}}=\frac{2\left( \frac{a}{4} \right)}{{{y}_{1}}}\,(x-{{x}_{1}})\]                    Þ \[y=\frac{a}{2{{y}_{1}}}(x)\,-\frac{a{{x}_{1}}}{2{{y}_{1}}}+{{y}_{1}}\]                    Here, \[m=\frac{a}{2{{y}_{1}}}=\,\tan \,{{45}^{o}}\]\[\Rightarrow \,\]\[\frac{a}{2{{y}_{1}}}=1\] Þ \[{{y}_{1}}=\frac{a}{2}\]                    From (i), \[{{x}_{1}}=\frac{a}{4}\].   \ Point is \[\left( \frac{a}{4},\,\frac{a}{2} \right)\].


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