Tangents are drawn from point (1, 1) to the e | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The equation of the ellipse is $x^2 + 4y^2 - 2x + 8y + 1 = 0$. Completing the square,we get $(x-1)^2 + 4(y+1)^2 = 4$,which simplifies to $\frac{(x

Vedclass · Accepted Answer

lies inside the ellipse $S = 0$

Vedclass · Answer

(A) The equation of the ellipse is $x^2 + 4y^2 - 2x + 8y + 1 = 0$. Completing the square,we get $(x-1)^2 + 4(y+1)^2 = 4$,which simplifies to $\frac{(x-1)^2}{4} + \frac{(y+1)^2}{1} = 1$. Let the tangent line passing through $(1, 1)$ be $y - 1 = m(x - 1)$,or $y = mx + (1 - m)$. For a line $y = mx + c$ to be a tangent to $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$,the condition is $(c - k + mh)^2 = a^2m^2 + b^2$. Here $h=1, k=-1, a^2=4, b^2=1$. So,$(1 - m - (-1) + m(1))^2 = 4m^2 + 1$. $(2)^2 = 4m^2 + 1 \Rightarrow 4 = 4m^2 + 1 \Rightarrow 4m^2 = 3 \Rightarrow m = \pm \frac{\sqrt{3}}{2}$. Given $m_1 > m_2$,we have $m_1 = \frac{\sqrt{3}}{2}$ and $m_2 = -\frac{\sqrt{3}}{2}$. The point $P$ is $(\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2})$. Substituting $P$ into the ellipse equation $S(x, y) = \frac{(x-1)^2}{4} + (y+1)^2 - 1$: $S(\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}) = \frac{(\frac{\sqrt{3}}{2} - 1)^2}{4} + (-\frac{\sqrt{3}}{2} + 1)^2 - 1$. Since $(\frac{\sqrt{3}}{2} - 1)^2 = (1 - \frac{\sqrt{3}}{2})^2$,we have $S = \frac{(1 - \frac{\sqrt{3}}{2})^2}{4} + (1 - \frac{\sqrt{3}}{2})^2 - 1 = \frac{5}{4}(1 - \frac{\sqrt{3}}{2})^2 - 1$. Since $(1 - \frac{\sqrt{3}}{2})^2 \approx (1 - 0.866)^2 = (0.134)^2 \approx 0.018$,$S \approx \frac{5}{4}(0.018) - 1 Thus,the point lies inside the ellipse.

Tangents are drawn from point $(1, 1)$ to the ellipse $S \equiv x^2 + 4y^2 - 2x + 8y + 1 = 0$. If $m_1, m_2$ $(m_1 > m_2)$ are the slopes of these tangents,then with respect to the given ellipse,the point $P(m_1, m_2)$:

Similar Questions

The eccentricity of the conic $16x^2 + 7y^2 = 112$ is

The area (in square units) of the quadrilateral formed by joining the foci of the two ellipses $\frac{x^2}{9}+\frac{y^2}{5}=1$ and $\frac{x^2}{5}+\frac{y^2}{9}=1$ is

The equation of the ellipse whose centre is $(2, -3)$,one of the foci is $(3, -3)$ and the corresponding vertex is $(4, -3)$ is

If $A_1, A_2, A_3$ are the areas of the ellipse $x^2+4y^2-4=0$, its director circle, and its auxiliary circle respectively, then $A_2+A_3-A_1=$ (in $\pi$)

If a normal is drawn at a variable point $P(x, y)$ on the curve $9x^2 + 16y^2 = 144$,then the maximum distance from the centre of the curve to the normal is

Vedclass Test Series

Exam Paper Generator

Online Exam Module