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दिये गये वक्र `y(x-2)(x-3)-x+7=0` के स्पर्शी व अभिलम्ब के समीकरण उस बिन्दु पर ज्ञात कीजिए जहाँ यह x-अक्ष को काटता है |

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Correct Answer - `20y=x-7, 20x+y=140`
दी गई समीकरण इस प्रकार लिखी जा सकती है,
`yx^(2)-5xy+6y-x+7=0`
y = 0 पर, x = 7 अभीष्ट बिन्दु (7, 0) समीकरण (1) से,
`(dy)/(dx)=(1+5y-2xy)/(x^(2)-5x+6)=(1)/(20), (7, 0)` पर
`:.` बिन्दु (7, 0) पर स्पर्श रेखा का समीकरण
`y-0=(1)/(20)(x-7)implies20y=x-7`
तथा अभिलम्ब का समीकरण
`y-0=-20(x-7)`
`implies" "20x+y=140`

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