The normal at the point (bt_1^2, 2bt_1) on a parabola meets the parabola again in the point (bt_2^2, 2bt_2) , then: - Mathematics MCQ AIEEE-CBSE-ENG-03 2026 | ExamTest.live

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The normal at the point (b t_{1}^{2}, 2 b t_{1}) on a parabola meets the parabola again in the point (b t_{2}^{2}, 2 b t_{2}), then:

Solution & Step-by-step Explanation

The equation of the normal to the parabola y^{2} = 4 b x at (b t_{1}^{2}, 2 b t_{1}) is: y = - t_{1} x + 2 b t_{1} + b t_{1}^{3} If this normal passes through (b t_{2}^{2}, 2 b t_{2}), we substitute the coordinates: 2 b t_{2} = - t_{1} (b t_{2}^{2}) + 2 b t_{1} + b t_{1}^{3} 2 b (t_{2} - t_{1}) = - b t_{1} (t_{2}^{2} - t_{1}^{2}) 2 (t_{2} - t_{1}) = - t_{1} (t_{2} - t_{1}) (t_{2} + t_{1}) Since t_{1} \neq = t_{2}, we divide by (t_{2} - t_{1}) : 2 = - t_{1} (t_{2} + t_{1}) - \frac{2}{t _{1}} = t_{2} + t_{1} ⟹ t_{2} = - t_{1} - \frac{2}{t _{1}}

Try it yourself before checking the explanation above.

The normal at the point (b t_{1}^{2}, 2 b t_{1}) on a parabola meets the parabola again in the point (b t_{2}^{2}, 2 b t_{2}), then:

t_{2} = - t_{1} - \frac{2}{t _{1}}

t_{2} = - t_{1} + \frac{2}{t _{1}}

t_{2} = t_{1} - \frac{2}{t _{1}}

t_{2} = t_{1} + \frac{2}{t _{1}}

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