If x+2y+k=0, k0 is a tangent to the ellipse | vedclass

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(A) Given the ellipse equation: $2x^2+y^2=2$,which can be written as $x^2+\frac{y^2}{2}=1$. Here,$a^2=1$ and $b^2=2$. The equation of the tangent is $

Vedclass · Accepted Answer

$\sqrt{2}x-2y+1=0$

Vedclass · Answer

(A) Given the ellipse equation: $2x^2+y^2=2$,which can be written as $x^2+\frac{y^2}{2}=1$. Here,$a^2=1$ and $b^2=2$. The equation of the tangent is $x+2y+k=0$,which implies $y=-\frac{1}{2}x-\frac{k}{2}$. Comparing this with $y=mx+c$,we get $m=-\frac{1}{2}$ and $c=-\frac{k}{2}$. The condition for a line $y=mx+c$ to be a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $c^2=a^2m^2+b^2$. Substituting the values: $(-\frac{k}{2})^2 = (1)(-\frac{1}{2})^2 + 2$. $\frac{k^2}{4} = \frac{1}{4} + 2 = \frac{9}{4}$. $k^2=9$,so $k=\pm 3$. Since $k>0$,we have $k=3$. The point of contact is $\left(\frac{1}{\sqrt{2}}, \frac{3}{3}\right) = \left(\frac{1}{\sqrt{2}}, 1\right)$. The equation of the normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $(x_1, y_1)$ is $\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2$. Substituting $a^2=1, b^2=2, x_1=\frac{1}{\sqrt{2}}, y_1=1$: $\frac{1 \cdot x}{1/\sqrt{2}} - \frac{2 \cdot y}{1} = 1-2$. $\sqrt{2}x - 2y = -1$,which simplifies to $\sqrt{2}x-2y+1=0$.

If $x+2y+k=0, k>0$ is a tangent to the ellipse $2x^2+y^2=2$,then the equation of the normal to the given ellipse at $\left(\frac{1}{\sqrt{2}}, \frac{k}{3}\right)$ is:

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