Question

If the tangent and the normal drawn to the curve $xy^2 + x^2y = 12$ at the point (1,3) meet the X-axis in T and N respectively, then TN =

• A. $7/5$
• B. $45/7$
• C. $3\sqrt{274}/7$
• D. $274/35$

Hint:

To find the x-intercept of a line, set $y=0$ in its equation. The distance between two points on the x-axis, $(x_1, 0)$ and $(x_2, 0)$, is simply $|x_1 - x_2|$.

Answer 1

The Correct Option is D

The curve is given by $xy^2 + x^2y = 12$. The point is $P(1,3)$.
Step 1: Find the slope of the tangent by implicit differentiation.
Differentiating with respect to $x$: $\frac{d}{dx}(xy^2) + \frac{d}{dx}(x^2y) = \frac{d}{dx}(12)$.
$(1 \cdot y^2 + x \cdot 2y \frac{dy}{dx}) + (2x \cdot y + x^2 \cdot \frac{dy}{dx}) = 0$.
$\frac{dy}{dx}(2xy+x^2) = -y^2-2xy$.
$\frac{dy}{dx} = -\frac{y^2+2xy}{x^2+2xy}$.
At the point $(1,3)$, the slope of the tangent ($m_T$) is:
$m_T = -\frac{3^2+2(1)(3)}{1^2+2(1)(3)} = -\frac{9+6}{1+6} = -\frac{15}{7}$.
Step 2: Find the equation of the tangent and the coordinate of T.
The equation of the tangent is $y-3 = -\frac{15}{7}(x-1)$.
To find where it meets the X-axis (point T), set $y=0$:
$0-3 = -\frac{15}{7}(x_T-1) \implies -21 = -15(x_T-1) \implies \frac{21}{15} = x_T-1 \implies x_T = 1+\frac{7}{5}=\frac{12}{5}$.
So, T is the point $(\frac{12}{5}, 0)$.
Step 3: Find the equation of the normal and the coordinate of N.
The slope of the normal ($m_N$) is the negative reciprocal of the tangent's slope:
$m_N = -\frac{1}{m_T} = -\frac{1}{-15/7} = \frac{7}{15}$.
The equation of the normal is $y-3 = \frac{7}{15}(x-1)$.
To find where it meets the X-axis (point N), set $y=0$:
$0-3 = \frac{7}{15}(x_N-1) \implies -45 = 7(x_N-1) \implies -45 = 7x_N - 7 \implies 7x_N = -38 \implies x_N = -\frac{38}{7}$.
So, N is the point $(-\frac{38}{7}, 0)$.
Step 4: Calculate the distance TN.
Since both points lie on the X-axis, the distance is the absolute difference of their x-coordinates.
$TN = |x_T - x_N| = \left|\frac{12}{5} - \left(-\frac{38}{7}\right)\right| = \left|\frac{12}{5} + \frac{38}{7}\right|$.
$TN = \left|\frac{12(7) + 38(5)}{35}\right| = \left|\frac{84 + 190}{35}\right| = \frac{274}{35}$.