A point (4, -1) lies on an ellipse whose axes | vedclass

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(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Since the point $(4, -1)$ lies on the ellipse,we have $\frac{16}{a^2}

Vedclass · Accepted Answer

$\frac{x^2}{20} + \frac{y^2}{5} = 1$

Vedclass · Answer

(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Since the point $(4, -1)$ lies on the ellipse,we have $\frac{16}{a^2} + \frac{1}{b^2} = 1$,which implies $16b^2 + a^2 = a^2b^2$. The equation of the tangent at $(x_1, y_1)$ to the ellipse is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$. Substituting $(4, -1)$,we get $\frac{4x}{a^2} - \frac{y}{b^2} = 1$. Comparing this with the given tangent $x + 4y = 10$,or $\frac{x}{10} + \frac{4y}{10} = 1$,we have: $\frac{4/a^2}{1/10} = \frac{-1/b^2}{4/10} = \frac{1}{1}$. Thus,$\frac{40}{a^2} = 1 \Rightarrow a^2 = 40$ and $\frac{-10}{4b^2} = 1 \Rightarrow b^2 = -2.5$. Wait,re-evaluating: The condition for $y = mx + c$ to be a tangent is $c^2 = a^2m^2 + b^2$. Given $x + 4y = 10 \Rightarrow y = -\frac{1}{4}x + \frac{5}{2}$,so $m = -\frac{1}{4}$ and $c = \frac{5}{2}$. Then $c^2 = a^2m^2 + b^2$ $\Rightarrow \frac{25}{4} = \frac{a^2}{16} + b^2$ $\Rightarrow 100 = a^2 + 16b^2$. Also,$(4, -1)$ on the ellipse gives $\frac{16}{a^2} + \frac{1}{b^2} = 1 \Rightarrow 16b^2 + a^2 = a^2b^2$. Substituting $100 = a^2b^2$,we get $a^2 + 16b^2 = 100$ and $a^2b^2 = 100$. Let $X = a^2, Y = b^2$. Then $X + 16Y = 100$ and $XY = 100 \Rightarrow Y = 100/X$. $X + 16(100/X) = 100 \Rightarrow X^2 - 100X + 1600 = 0$. $(X - 80)(X - 20) = 0$. If $a^2 = 20$,then $b^2 = 5$. If $a^2 = 80$,then $b^2 = 1.25$. Checking the options,$\frac{x^2}{20} + \frac{y^2}{5} = 1$ matches option $C$.

$A$ point $(4, -1)$ lies on an ellipse whose axes are parallel to the coordinate axes. If the line $x + 4y - 10 = 0$ is tangent to the ellipse at this point,find its equation $(a > b)$.

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