SATHEE: Ellipse 2 Question 4

Ellipse 2 Question 4

4. If the tangent to the parabola $y^{2} = x$ at a point $(α, β), (β > 0)$ is also a tangent to the ellipse, $x^{2} + 2 y^{2} = 1$ , then $α$ is equal to

(2019 Main, 9 April II)

(a) $\sqrt{2} + 1$

(b) $\sqrt{2} - 1$

(d) $2 \sqrt{2} - 1$

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Solution:

Since the point $(α, β)$ is on the parabola $y^{2} = x$ , so

$α = β^{2}$

Now, equation of tangent at point $(α, β)$ to the parabola $y^{2} = x$ , is $T = 0$

$\Rightarrow y β = \frac{1}{2} (x + α)$

$[∵$ equation of the tangent to the parabola $y^{2} = 4 a x$ at a point $(x_{1}, y_{1})$ is given by $y y_{1} = 2 a (x + x_{1})]$

$\begin{array}{ll} \Rightarrow & 2 y β = x + β^{2} \\ \Rightarrow & y = \frac{x}{2 β} + \frac{β}{2} \end{array}$

Since, line (ii) is also a tangent of the ellipse

$x^{2} + 2 y^{2} = 1$

$[∵$ condition of tangency of line $y = m x + c$ to ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is $c^{2} = a^{2} m^{2} + b^{2}$ ,

here

$m = \frac{1}{2 β}, a = 1, b = \frac{1}{\sqrt{2}} and c = \frac{β}{2}$

$\Rightarrow \frac{β^{2}}{4} = \frac{1}{4 β^{2}} + \frac{1}{\sqrt{2}}$

$\Rightarrow β^{4} = 1 + 2 β^{2}$

$\Rightarrow β^{4} - 2 β^{2} - 1 = 0$

$\Rightarrow β^{2} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2 \sqrt{2}}{2} = 1 \pm \sqrt{2}$

$\Rightarrow β^{2} = 1 + \sqrt{2}$

$∵ α = β^{2} = 1 + \sqrt{2}$

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Ellipse 2 Question 4

4. If the tangent to the parabola $y^{2} = x$ at a point $(α, β), (β > 0)$ is also a tangent to the ellipse, $x^{2} + 2 y^{2} = 1$ , then $α$ is equal to

Solution:

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Ellipse 2 Question 4

4. If the tangent to the parabola y2=x at a point (α,β),(β>0) is also a tangent to the ellipse, x2+2y2=1, then α is equal to

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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4. If the tangent to the parabola $y^{2} = x$ at a point $(α, β), (β > 0)$ is also a tangent to the ellipse, $x^{2} + 2 y^{2} = 1$ , then $α$ is equal to