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Q. The equation of the hyperbola whose foci are (–2,0) and (2, 0) and eccentricity is 2 is given by :(1) $-3 x^{2}+y^{2}=3$(2) $x^{2}-3 y^{2}=3$(3) $3 x^{2}-y^{2}=3$(4) $-x^{2}+3 y^{2}=3$ [AIEEE-2011]
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Sol. (3)$2 \mathrm{ae}=4$$\mathrm{ae}=2$$\mathrm{a}(2)=2$$\mathrm{a}=1$$\mathrm{b}^{2}=\mathrm{a}^{2}\left(\mathrm{e}^{2}-1\right)$$\begin{aligned}=& 1(4-1)=3 \\ \text { equation } & \frac{\mathrm{x}^{2}}{1}-\frac{\mathrm{y}^{2}}{3}=1 \\ & 3 \mathrm{x}^{2}-\mathrm{y}^{2}=3 \end{aligned}$
Q. A tangent to the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$ meets $x$ -axis at $P$ and $y$ -axis at Q. Lines PR and QR are drawn such that OPRQ is a rectangle (where O is the origin). Then R lies on :(1) $\frac{2}{x^{2}}-\frac{4}{y^{2}}=1$(2) $\frac{4}{x^{2}}-\frac{2}{y^{2}}=1$(3) $\frac{4}{x^{2}}+\frac{2}{y^{2}}=1$(4) $\frac{2}{x^{2}}+\frac{4}{y^{2}}=1$ [JEE-Main (On line)-2013]
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Sol. (2)Tangent is $\frac{x}{2} \sec \theta-\frac{y}{\sqrt{2}} \tan \theta=1$$P\left(\frac{2}{\sec \theta}, 0\right), Q\left(0,-\frac{\sqrt{2}}{\tan \theta}\right)$$R\left(\frac{2}{\sec \theta},-\frac{\sqrt{2}}{\tan \theta}\right)=(h, k)$Locus of $R \Rightarrow \quad \frac{4}{h^{2}}-\frac{9}{k^{2}}=1=\frac{4}{x^{2}}-\frac{9}{y^{2}}=1$
Q. A common tangent to the conics $x^{2}=6 y$ and $2 x^{2}-4 y^{2}=9$ is :(1) $x+y=\frac{9}{2}$(2) $x+y=1$(3) $x-y=\frac{3}{2}$(4) x – y = 1 [JEE-Main (On line)-2013]
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Sol. (3)Equation of tangent for $\frac{x^{2}}{(9 / 2)}-\frac{y^{2}}{(9 / 4)}=1$is $y=m x \pm \sqrt{\frac{9}{2} m^{2}-\frac{9}{4}}$and tangent to $x^{2}=6 y$ is $y=m x-\frac{3}{2} m^{2}$ ….(2)Now equated ( 1) and ( 2)
Q. The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is :(1) $\sqrt{3}$(2) $\frac{4}{3}$(3) $\frac{4}{\sqrt{3}}$(4) $\frac{2}{\sqrt{3}}$ [JEE-Main 2016]
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Sol. (4)Given$\frac{2 \mathrm{b}^{2}}{\mathrm{a}}=8$$2 \mathrm{b}=\mathrm{ae}$we know$\mathrm{b}^{2}=\mathrm{a}^{2}\left(\mathrm{e}^{2}-1\right)$substitute $\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{e}}{2}$ from ( 2) in ( 3)$\Rightarrow \frac{\mathrm{e}^{2}}{4}=\mathrm{e}^{2}-1 \Rightarrow 4=3 \mathrm{e}^{2}$$\Rightarrow \mathrm{e}=\frac{2}{\sqrt{3}}$
Q. A hyperbola passes through the point $\mathrm{P}(\sqrt{2}, \sqrt{3})$ and has foci at $(\pm 2,0) .$ Then the tangent to this hyperbola at $\mathrm{P}$ also passes through the point:(1) $(-\sqrt{2},-\sqrt{3})$(2) $(3 \sqrt{2}, 2 \sqrt{3})$(3) $(2 \sqrt{2}, 3 \sqrt{3})$(4) $(\sqrt{3}, \sqrt{2})$ [JEE-Main 2017]
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Sol. (3)Equation of hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$foci is $(\pm 2,0)$ hence ae $=2, \Rightarrow a^{2} e^{2}=4$$b^{2}=a^{2}\left(e^{2}-1\right)$$\therefore a^{2}+b^{2}=4$ …(1)Hyperbola passes through $(\sqrt{2}, \sqrt{3})$$\therefore \frac{2}{\mathrm{a}^{2}}-\frac{3}{\mathrm{b}^{2}}=1$ …(2)On solving ( 1) and ( 2)$\mathrm{a}^{2}=8(\text { is rejected })$ and $\mathrm{a}^{2}=1$ and $\mathrm{b}^{2}=3$$\therefore \frac{\mathrm{x}^{2}}{1}-\frac{\mathrm{y}^{2}}{3}=1$Equation of tangent is $\frac{\sqrt{2} \mathrm{x}}{1}-\frac{\sqrt{3} \mathrm{y}}{3}=1$Hence $(2 \sqrt{2}, 3 \sqrt{3})$ satisfy it.
Q. Tangents are drawn to the hyperbola $4 \mathrm{x}^{2}-\mathrm{y}^{2}=36$ at the point $\mathrm{P}$ and $\mathrm{Q}$. If these tangents intersect at the point $\mathrm{T}(0,3)$ then the area (in sq. units) of $\Delta \mathrm{PTQ}$ is –(1) $54 \sqrt{3}$(2) $60 \sqrt{3}$(3) $36 \sqrt{5}$(4) $45 \sqrt{5}$ [JEE-Main 2018]
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Sol. (4)Equation PQ : chord of contact T = 0y = – 12
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solutions are not at all clear please try to give clear solutions
It’s not enough for us?
Correct ?
Right it is not enough we are doing more questions
thanks for the precious questions 🙂
🤟🤟🤟
Emmett dddddddddd Jenna am keen is add we nnaa all manner nee d see we let me lengthens meet meeru key kk lily
To GM
Interesting and also helpful to all
Plz give us some more problems
𝕐𝕖𝕤 𝕥𝕙𝕚𝕤 𝕚𝕤 𝕣𝕖𝕒𝕝𝕪 𝕙𝕖𝕝𝕡𝕗𝕦𝕝 𝕗𝕠𝕣 𝕒𝕟𝕪 𝕒𝕤𝕡𝕣𝕚𝕟𝕥 𝕨𝕙𝕠 𝕨𝕒𝕟𝕥 𝕥𝕠 𝕕𝕠 𝕓𝕖𝕥𝕥𝕖𝕣 𝕤𝕔𝕠𝕣𝕖 𝕚𝕟 𝕖𝕟𝕥𝕣𝕒𝕟𝕔𝕖 𝕖𝕩𝕒𝕞
Do u have some more questions ?…
No,these are not helpful
Chill bro
Solution of 4 question is not understood
i’m getting multiple answers in Q4
HI BRO
WE DONT NEED TO SEE THE SOLUTION FOR QUESTION NUMBER-4
ITS A PRETTY SIMPLE QUESTION
TRY ONCE MORE U WILL DEFINETLY GET IT