Question

5: P is a point on the parabola whose ordinate

equals its abscissa. A normal is drawn to the parabola at

P to meet it again at Q. If S is the focus of the parabola

then the product of the slopes of SP and SQ is-

Answer 1

The answer to your question is provided in the image:

$\text{Since the abscissa and ordinate are equal for point P}\\ 2at_1=at_1^2\\ t_1=2\\ \text{Slope PQ}\\ \dfrac{2at_1-2at_2}{at_1^2-at_2^2}=\dfrac{2}{t_1+t_2}\\ \dfrac{dy}{dx}=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}=\dfrac{1}{t_1}=\dfrac{1}{2}\\ \text{Product of slope PQ and tangent is -1}\\ \dfrac{2}{t_1+t_2}\times{\dfrac{1}{2}}=-1\\ t_1+t_2=-1\\ \therefore\\ t_2=-3\\ \text{As we know slope of line between two points is deter mined by }\\ {y_1-y_2\over x_1-x_2}\\ \text{Slope SP}\\ \dfrac{0-2at_1}{a-at_1^2}=\dfrac{4}{3}\\ \text{Slope SQ}\\ \dfrac{0-2at_2}{a-at_2^2}=\dfrac{-3}{4}\\ \therefore\\ \text{The product of slope SQ and slope SP is -1}$