दीर्घवृत्त `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`के बिन्दु `(x_(1), y_(1))` पर स्पर्श रेखा का समीकरण ज्ञात कीजिए.

Explanation

यहाँ `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1" "...(1)`
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
`(2x)/(a^(2))+ (2y)/(b^(2))(dy)/(dx) =0`
`implies (dy)/(dx) =-(b^(2) x)/(a^(2)y)`
`implies ((dy)/(dx))_("(x_(1), y_(1))). (x-y_(1))`
`impliesy-y_(1)=-(b^(2)x_(1))/(a^(2)y_(1))(x-x_(1))`
`implies (yy_(1)-y_(1)^(2))/(b^(2))=-((x x_(1)-x_(1)^(2))/(a^(2)))`
`implies (yy_(1)-y_(1)^(2))/(b^(2))-(y_(1)^(2))/(b^(2))=-(x x_(1))/(a^(2))+ (x_(1)^(2))/(a^(2))`
`implies (x x_(1))/(a^(2))+(yy_(1))/(b^(2))=1 `
`[because (x_(1), y_(1))` वक्र (1 ) पर स्थित है इसलिए `(x_(1)^(2))/(a^(2))+ (y_(1)^(2))/(b^(2))=1"]"`
बिन्दु `(x_(1), y_(1))` पर अभिलम्ब का समीकरण है-
`implies y-y_(1) =-(1)/(((dy)/(dx))_(""(x_(1), y_(1))))(x-x_(1))`
`implies y-y_(1) =(-1)/(-(b^(2)x_(1))/(a^(2)y_(1)))(x-x_(1))`
`implies y-y_(1)=(a^(2)y_(1))/(b^(2)x_(1))(x-x_(1))`
`implies(b^(2) (y-y_(1)))/(y_(1))=(a^(2)(x-x_(1)))/(x_(1))`
`implies (b^(2)y)/(y_(1))=b^(2)=(a^(2) x)/(x_(1))-a^(2)`
`implies(a^(2)x)/(x_(1))-(b^(2) y)/(y_(1))=a^(2)-b^(2).`