Question

Find the point of contact when (i) Equation of parabola is $y^2 = 8x$ and slope of tangent is 2. (ii) Equation of parabola is $y^2 = -2x$ and slope of tangent is -2. (iii) Equation of parabola is $x^2 = -8y$ and slope of tangent is 3. (iv) Equation of parabola is $y^2 = 4(1 - x)$ and slope of tangent is -1.

• A. (i) $(\frac{1}{2}, 2)$; (ii) $(-\frac{1}{8}, \frac{1}{2})$; (iii) $(-\frac{4}{3}, 18)$; (iv) $(0, 2)$
• B. (i) $(\frac{1}{2}, 2)$; (ii) $(\frac{1}{8}, \frac{1}{2})$; (iii) $(\frac{4}{3}, 18)$; (iv) $(0, 2)$
• C. (i) $(2, \frac{1}{2})$; (ii) $(-\frac{1}{8}, \frac{1}{2})$; (iii) $(-\frac{4}{3}, 18)$; (iv) $(2, 0)$
• D. (i) $(\frac{1}{2}, 2)$; (ii) $(-\frac{1}{8}, -\frac{1}{2})$; (iii) $(\frac{4}{3}, 18)$; (iv) $(0, 2)$

Answer 1

Answer: (A)

(i) $(\frac{1}{2}, 2)$; (ii) $(-\frac{1}{8}, \frac{1}{2})$; (iii) $(-\frac{4}{3}, 18)$; (iv) $(0, 2)

Answer 2

(i) For a parabola $y^2 = 4ax$, the point of contact of a tangent with slope $m$ is given by $(\frac{a}{m^2}, \frac{2a}{m})$. The given parabola is $y^2 = 8x$. Comparing with $y^2 = 4ax$, we get $4a = 8$, so $a = 2$. The slope of the tangent is given as $m = 2$. The point of contact is $(\frac{a}{m^2}, \frac{2a}{m}) = (\frac{2}{2^2}, \frac{2 \times 2}{2}) = (\frac{2}{4}, \frac{4}{2}) = (\frac{1}{2}, 2)$. (ii) The given parabola is $y^2 = -2x$. Comparing with $y^2 = 4ax$, we get $4a = -2$, so $a = -\frac{1}{2}$. The slope of the tangent is given as $m = -2$. The point of contact is $(\frac{a}{m^2}, \frac{2a}{m}) = (\frac{-1/2}{(-2)^2}, \frac{2 \times (-1/2)}{-2}) = (\frac{-1/2}{4}, \frac{-1}{-2}) = (-\frac{1}{8}, \frac{1}{2})$. (iii) For a parabola $x^2 = 4ay$, the point of contact of a tangent with slope $m$ is given by $(\frac{2a}{m}, -am^2)$. The given parabola is $x^2 = -8y$. Comparing with $x^2 = 4ay$, we get $4a = -8$, so $a = -2$. The slope of the tangent is given as $m = 3$. The point of contact is $(\frac{2a}{m}, -am^2) = (\frac{2 \times (-2)}{3}, -(-2) \times 3^2) = (-\frac{4}{3}, -(-2) \times 9) = (-\frac{4}{3}, 18)$. (iv) The parabola is $y^2 = 4(1 - x) \implies y^2 = -4(x - 1)$. Let $X = x-1$ and $Y = y$. The equation becomes $Y^2 = -4X$. This is of the form $Y^2 = 4a'X$, with $4a' = -4$, so $a' = -1$. The point of contact for $Y^2 = 4a'X$ with slope $m$ is $(\frac{a'}{m^2}, \frac{2a'}{m})$. Given slope $m = -1$. Point of contact in $(X, Y)$ coordinates: $(\frac{-1}{(-1)^2}, \frac{2 \times (-1)}{-1}) = (\frac{-1}{1}, \frac{-2}{-1}) = (-1, 2)$. Now, convert back to $(x, y)$ coordinates: $X = x-1 \implies -1 = x-1 \implies x = 0$. $Y = y \implies y = 2$. The point of contact is $(0, 2)$.