SATHEE: Parabola 2 Question 1

Parabola 2 Question 1

1. If the line $a x + y = c$ , touches both the curves $x^{2} + y^{2} = 1$ and $y^{2} = 4 \sqrt{2} x$ , then $| c |$ is equal to

(2019 Main, 10 April, II)

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

(b) 2

(d) $\frac{1}{2}$

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Answer:

Correct Answer: 1. (c)

Solution:

Key Idea Use the equation of tangent of slope ’ $m$ ’ to the parabola $y^{2} = 4 a x$ is $y = m x + \frac{a}{m}$ and a line $a x + b y + c = 0$ touches the circle $x^{2} + y^{2} = r^{2}$ , if $\frac{| c |}{\sqrt{a^{2} + b^{2}}} = r$ .

Since, equation of given parabola is $y^{2} = 4 \sqrt{2} x$ and equation of tangent line is $a x + y = c$ or $y = - a x + c$ ,

then $c = \frac{\sqrt{2}}{m} = \frac{\sqrt{2}}{- a} [∵ m =$ slope of line $= - a]$

$[∵$ line $y = m x + c$ touches the parabola

$y^{2} = 4 a x$ iff $c = a / m$ ]

Then, equation of tangent line becomes

$y = - a x - \frac{\sqrt{2}}{a}$

$∵$ Line (i) is also tangent to the circle $x^{2} + y^{2} = 1$ .

$∴$ Radius $= 1 = \frac{| - \frac{\sqrt{2}}{a} |}{\sqrt{1 + a^{2}}} \Rightarrow \sqrt{1 + a^{2}} = | - \frac{\sqrt{2}}{a} |$

$\Rightarrow 1 + a^{2} = \frac{2}{a^{2}} [squaring both sides]$

$\Rightarrow a^{4} + a^{2} - 2 = 0 \Rightarrow (a^{2} + 2) (a^{2} - 1) = 0$

$\Rightarrow a^{2} = 1 [∵ a^{2} > 0, \forall a \in R]$

$∴ | c | = \frac{\sqrt{2}}{| a |} = \sqrt{2}$

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Parabola 2 Question 1

1. If the line $a x + y = c$ , touches both the curves $x^{2} + y^{2} = 1$ and $y^{2} = 4 \sqrt{2} x$ , then $| c |$ is equal to

Answer:

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Parabola 2 Question 1

1. If the line ax+y=c, touches both the curves x2+y2=1 and y2=42x, then |c| is equal to

Answer:

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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1. If the line $a x + y = c$ , touches both the curves $x^{2} + y^{2} = 1$ and $y^{2} = 4 \sqrt{2} x$ , then $| c |$ is equal to