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Q.Show that the line x−y+4=0x-y+4=0 is a tangent to the ellipse x2+3y2=12x^2+3y^2=12. Find the point of contact.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2016Subjective· 10mImportance★★★★★
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Verify the ellipse-tangency condition c2=a2m2+b2c^2=a^2m^2+b^2 for the line x−y+4=0x-y+4=0 on x2+3y2=12x^2+3y^2=12, then locate the point of contact.

  1. Standard form of the ellipse. Divide x2+3y2=12x^2+3y^2=12 by 1212: x212+y24=1\dfrac{x^2}{12}+\dfrac{y^2}{4}=1. So a2=12a^2=12, b2=4b^2=4.

  2. Standard form of the line. x−y+4=0⇒y=x+4x-y+4=0 \Rightarrow y=x+4, which is of the form y=mx+cy=mx+c with m=1m=1, c=4c=4.

  3. Tangency condition. A line y=mx+cy=mx+c touches the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 if and only if c2=a2m2+b2c^2=a^2m^2+b^2.

    Compute the right side: a2m2+b2=12(1)2+4=12+4=16a^2m^2+b^2=12(1)^2+4=12+4=16.

    Compute the left side: c2=42=16c^2=4^2=16.

    Since 16=1616=16, the condition holds, so the line x−y+4=0x-y+4=0 is indeed a tangent to the ellipse.

    …

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