Question

Find the equation of tangent(s) to the parabola

• A. y^2 = 3x at the point (3,-3)
• B. y^2 = 2x at the point (2,-2)
• C. x^2 = 4y at the point (-4,4)
• D. y^2 = 2x whose slope is 5/4
• E. y^2 = 8x which is perpendicular to x + 2y + 7 = 0
• F. x^2 = -2y which is parallel to 4x + y + 1 = 0
• G. y^2 = 12x, which passes through the point (2,5)
• H. y^2 = 2x at the ends of latus rectum of parabola.
• I. y^2 = 8x and making an angle of 45^\circ with the line y = 3x + 5.
• J. x^2 = 2y, which passes through (1,-1).

Answer 1

Answer: (A), (B), (C), (D), (E), (F), (G), (H), (I), (J)

(i) $x + 2y + 3 = 0$ (ii) $x + 2y + 2 = 0$ (iii) $2x + y + 4 = 0$ (iv) $25x - 20y + 8 = 0$ (v) $2x - y + 1 = 0$ (vi) $4x + y - 8 = 0$ (vii) $x - y + 3 = 0$ and $3x - 2y + 4 = 0$ (viii) $2x - 2y + 1 = 0$ and $2x + 2y + 1 = 0$ (ix) $2x + y + 1 = 0$ and $x - 2y + 8 = 0$ (x) $(1 + \sqrt{3})x - y - (2 + \sqrt{3}) = 0$ and $(1 - \sqrt{3})x - y - (2 - \sqrt{3}) = 0$

Answer 2

(i) For $y^2 = 3x$, $4a = 3 \implies a = \frac{3}{4}$. Tangent at $(x_1, y_1)$ is $yy_1 = 2a(x+x_1)$. At $(3, -3)$: $y(-3) = 2\left(\frac{3}{4}\right)(x+3) \implies -3y = \frac{3}{2}(x+3) \implies -6y = 3x+9 \implies 3x+6y+9=0 \implies x+2y+3=0$.

(ii) For $y^2 = 2x$, $4a = 2 \implies a = \frac{1}{2}$. Tangent at $(2, -2)$: $y(-2) = 2\left(\frac{1}{2}\right)(x+2) \implies -2y = x+2 \implies x+2y+2=0$.

(iii) For $x^2 = 4y$, $4a = 4 \implies a = 1$. Tangent at $(-4, 4)$: $xx_1 = 2a(y+y_1) \implies x(-4) = 2(1)(y+4) \implies -4x = 2y+8 \implies 4x+2y+8=0 \implies 2x+y+4=0$.

(iv) For $y^2 = 2x$, $4a = 2 \implies a = \frac{1}{2}$. Tangent with slope $m = \frac{5}{4}$ is $y = mx + \frac{a}{m}$. $y = \frac{5}{4}x + \frac{1/2}{5/4} = \frac{5}{4}x + \frac{2}{5} \implies 20y = 25x + 8 \implies 25x - 20y + 8 = 0$.

(v) For $y^2 = 8x$, $4a = 8 \implies a = 2$. Line $x+2y+7=0$ has slope $-\frac{1}{2}$. Perpendicular tangent has slope $m = 2$. Tangent equation: $y = mx + \frac{a}{m} = 2x + \frac{2}{2} = 2x+1 \implies 2x-y+1=0$.

(vi) For $x^2 = -2y$, $-4a = -2 \implies a = \frac{1}{2}$. Line $4x+y+1=0$ has slope $-4$. Parallel tangent has slope $m = -4$. Tangent equation for $x^2 = -4ay$ is $y = mx + am^2$. $y = -4x + \frac{1}{2}(-4)^2 = -4x + 8 \implies 4x+y-8=0$.

(vii) For $y^2 = 12x$, $4a = 12 \implies a = 3$. Tangent through $(2,5)$ is $y = mx + \frac{a}{m}$. $5 = 2m + \frac{3}{m} \implies 5m = 2m^2 + 3 \implies 2m^2 - 5m + 3 = 0 \implies (2m-3)(m-1) = 0$. $m=1$ or $m=\frac{3}{2}$. If $m=1$, $y = x + 3 \implies x-y+3=0$. If $m=\frac{3}{2}$, $y = \frac{3}{2}x + \frac{3}{3/2} = \frac{3}{2}x + 2 \implies 2y = 3x+4 \implies 3x-2y+4=0$.

(viii) For $y^2 = 2x$, $4a = 2 \implies a = \frac{1}{2}$. Ends of latus rectum are $(a, \pm 2a) = (\frac{1}{2}, \pm 1)$. At $(\frac{1}{2}, 1)$: $y(1) = 2\left(\frac{1}{2}\right)(x+\frac{1}{2}) \implies y = x+\frac{1}{2} \implies 2x-2y+1=0$. At $(\frac{1}{2}, -1)$: $y(-1) = 2\left(\frac{1}{2}\right)(x+\frac{1}{2}) \implies -y = x+\frac{1}{2} \implies 2x+2y+1=0$.

(ix) For $y^2 = 8x$, $4a = 8 \implies a = 2$. Line $y=3x+5$ has slope $m_2=3$. Angle $45^\circ$. $\tan 45^\circ = |\frac{m-3}{1+3m}| = 1$. $\frac{m-3}{1+3m} = 1 \implies m-3 = 1+3m \implies 2m = -4 \implies m = -2$. $\frac{m-3}{1+3m} = -1 \implies m-3 = -1-3m \implies 4m = 2 \implies m = \frac{1}{2}$. Tangent equation $y = mx + \frac{a}{m}$. For $m=-2$, $y = -2x + \frac{2}{-2} = -2x-1 \implies 2x+y+1=0$. For $m=\frac{1}{2}$, $y = \frac{1}{2}x + \frac{2}{1/2} = \frac{1}{2}x+4 \implies x-2y+8=0$.

(x) For $x^2 = 2y$, $4a = 2 \implies a = \frac{1}{2}$. Tangent through $(1,-1)$. Tangent equation $y = mx - am^2$. Point $(1,-1)$ is on the tangent: $-1 = m(1) - am^2 \implies -1 = m - \frac{1}{2}m^2 \implies m^2 - 2m - 2 = 0$. $m = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$. For $m = 1+\sqrt{3}$: $y = (1+\sqrt{3})x - \frac{1}{2}(1+\sqrt{3})^2 = (1+\sqrt{3})x - \frac{1}{2}(1+2\sqrt{3}+3) = (1+\sqrt{3})x - (2+\sqrt{3})$. $(1+\sqrt{3})x - y - (2+\sqrt{3}) = 0$. For $m = 1-\sqrt{3}$: $y = (1-\sqrt{3})x - \frac{1}{2}(1-\sqrt{3})^2 = (1-\sqrt{3})x - \frac{1}{2}(1-2\sqrt{3}+3) = (1-\sqrt{3})x - (2-\sqrt{3})$. $(1-\sqrt{3})x - y - (2-\sqrt{3}) = 0$.